Mixing
Groups of order $2p$
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
There are a few question on classification of groups of order $2p$ on MSE but I'd like to receive a feedback on this proof (and have a question about it at the end).
Let $G$ be a group of order $2p$. If there is an element of order $2p$ in $G$, then the group is cyclic. Suppose there is no such element. We claim that $Gsimeq D_p=<a,b| a^p=1,b^2=1, baba=1>$.
By Cauchy's theorem, $G$ contains an element $a$ of order $p$. Then it contains its cyclic subgroup $H=1,a,dots,a^p-1$. Let $b$ be an element outside of this subgroup. The $p$ elements $b, ba,dots ba^p-1$ are pairwise different (if $ba^k=ba^m$ for $0le k,m < p$, then $a^k=a^m$, which contradicts to the fact that $H$ contains pairwise different elements). Moreover for no $k$ and $m$ as above can the equality $a^k=ba^m$ hold (if it holds, then $b=a^ka^-m$, which contradicts that $bnotin H$). Thus $G=Hcup bH$ as a set.
We proceed to prove that $b^2=1$. It cannot be the case that $b^2in H-1$ because if $b^2=a^r$ for $0 < r < p$ then $a=(a^r)^l=(b^2)^limplies b=a^-2l$, which contradicts to the fact that $bnotin H$ (here we use that $H$ can be generated by any non-identity element because $p$ is prime; in particular, $H=<a^r>$ so that $a=(a^r)^l$ for $l$ an integer). We cannot have $b^2in bH$ either because $b^2=ba^k$ yields $bin H$. Thus $b^2=1$.
Also $(ba)^2=1$. Indeed, if $babain bH$, then similarly $bin H$. If $babain H-1$, then $baba=a^rimplies bab=a^r-1implies b^2=a^r-2$, and as above, $a=(a^r-2)^l=(b^2)^limplies b=a^-2l$.
So $G$ is generated by $a,b$ and the relations $a^p=b^2=baba=1$ hold in $G$.
Is it sufficient to conclude that $G$ is isomorphic to $D_p$? Or do I need to verify somehow that there are no relations in $G$ other than those implied by the ones above? And are there any other gaps or mistakes in my proof?
abstract-algebra group-theory group-isomorphism group-presentation dihedral-groups
asked Aug 6 at 17:24
user437309
556212
add a comment |Â
up vote
3
down vote
favorite
There are a few question on classification of groups of order $2p$ on MSE but I'd like to receive a feedback on this proof (and have a question about it at the end).
Let $G$ be a group of order $2p$. If there is an element of order $2p$ in $G$, then the group is cyclic. Suppose there is no such element. We claim that $Gsimeq D_p=<a,b| a^p=1,b^2=1, baba=1>$.
By Cauchy's theorem, $G$ contains an element $a$ of order $p$. Then it contains its cyclic subgroup $H=1,a,dots,a^p-1$. Let $b$ be an element outside of this subgroup. The $p$ elements $b, ba,dots ba^p-1$ are pairwise different (if $ba^k=ba^m$ for $0le k,m < p$, then $a^k=a^m$, which contradicts to the fact that $H$ contains pairwise different elements). Moreover for no $k$ and $m$ as above can the equality $a^k=ba^m$ hold (if it holds, then $b=a^ka^-m$, which contradicts that $bnotin H$). Thus $G=Hcup bH$ as a set.
We proceed to prove that $b^2=1$. It cannot be the case that $b^2in H-1$ because if $b^2=a^r$ for $0 < r < p$ then $a=(a^r)^l=(b^2)^limplies b=a^-2l$, which contradicts to the fact that $bnotin H$ (here we use that $H$ can be generated by any non-identity element because $p$ is prime; in particular, $H=<a^r>$ so that $a=(a^r)^l$ for $l$ an integer). We cannot have $b^2in bH$ either because $b^2=ba^k$ yields $bin H$. Thus $b^2=1$.
Also $(ba)^2=1$. Indeed, if $babain bH$, then similarly $bin H$. If $babain H-1$, then $baba=a^rimplies bab=a^r-1implies b^2=a^r-2$, and as above, $a=(a^r-2)^l=(b^2)^limplies b=a^-2l$.
So $G$ is generated by $a,b$ and the relations $a^p=b^2=baba=1$ hold in $G$.
Is it sufficient to conclude that $G$ is isomorphic to $D_p$? Or do I need to verify somehow that there are no relations in $G$ other than those implied by the ones above? And are there any other gaps or mistakes in my proof?
abstract-algebra group-theory group-isomorphism group-presentation dihedral-groups
asked Aug 6 at 17:24
user437309
556212
1
How are you getting from $b^2l=a$ to $b=a^-2l$?
– Steve D
Aug 6 at 17:32
1
Here is one way to show $b^2neq a^r$: then $b$ would have order $2p$, and your group would be cyclic.
– Steve D
Aug 6 at 17:34
@SteveD Oops that's indeed wrong, thanks.
– user437309
Aug 6 at 17:36
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
There are a few question on classification of groups of order $2p$ on MSE but I'd like to receive a feedback on this proof (and have a question about it at the end).
Let $G$ be a group of order $2p$. If there is an element of order $2p$ in $G$, then the group is cyclic. Suppose there is no such element. We claim that $Gsimeq D_p=<a,b| a^p=1,b^2=1, baba=1>$.
By Cauchy's theorem, $G$ contains an element $a$ of order $p$. Then it contains its cyclic subgroup $H=1,a,dots,a^p-1$. Let $b$ be an element outside of this subgroup. The $p$ elements $b, ba,dots ba^p-1$ are pairwise different (if $ba^k=ba^m$ for $0le k,m < p$, then $a^k=a^m$, which contradicts to the fact that $H$ contains pairwise different elements). Moreover for no $k$ and $m$ as above can the equality $a^k=ba^m$ hold (if it holds, then $b=a^ka^-m$, which contradicts that $bnotin H$). Thus $G=Hcup bH$ as a set.
We proceed to prove that $b^2=1$. It cannot be the case that $b^2in H-1$ because if $b^2=a^r$ for $0 < r < p$ then $a=(a^r)^l=(b^2)^limplies b=a^-2l$, which contradicts to the fact that $bnotin H$ (here we use that $H$ can be generated by any non-identity element because $p$ is prime; in particular, $H=<a^r>$ so that $a=(a^r)^l$ for $l$ an integer). We cannot have $b^2in bH$ either because $b^2=ba^k$ yields $bin H$. Thus $b^2=1$.
Also $(ba)^2=1$. Indeed, if $babain bH$, then similarly $bin H$. If $babain H-1$, then $baba=a^rimplies bab=a^r-1implies b^2=a^r-2$, and as above, $a=(a^r-2)^l=(b^2)^limplies b=a^-2l$.
So $G$ is generated by $a,b$ and the relations $a^p=b^2=baba=1$ hold in $G$.
Is it sufficient to conclude that $G$ is isomorphic to $D_p$? Or do I need to verify somehow that there are no relations in $G$ other than those implied by the ones above? And are there any other gaps or mistakes in my proof?
abstract-algebra group-theory group-isomorphism group-presentation dihedral-groups
asked Aug 6 at 17:24
user437309
556212
There are a few question on classification of groups of order $2p$ on MSE but I'd like to receive a feedback on this proof (and have a question about it at the end).
Let $G$ be a group of order $2p$. If there is an element of order $2p$ in $G$, then the group is cyclic. Suppose there is no such element. We claim that $Gsimeq D_p=<a,b| a^p=1,b^2=1, baba=1>$.
By Cauchy's theorem, $G$ contains an element $a$ of order $p$. Then it contains its cyclic subgroup $H=1,a,dots,a^p-1$. Let $b$ be an element outside of this subgroup. The $p$ elements $b, ba,dots ba^p-1$ are pairwise different (if $ba^k=ba^m$ for $0le k,m < p$, then $a^k=a^m$, which contradicts to the fact that $H$ contains pairwise different elements). Moreover for no $k$ and $m$ as above can the equality $a^k=ba^m$ hold (if it holds, then $b=a^ka^-m$, which contradicts that $bnotin H$). Thus $G=Hcup bH$ as a set.
We proceed to prove that $b^2=1$. It cannot be the case that $b^2in H-1$ because if $b^2=a^r$ for $0 < r < p$ then $a=(a^r)^l=(b^2)^limplies b=a^-2l$, which contradicts to the fact that $bnotin H$ (here we use that $H$ can be generated by any non-identity element because $p$ is prime; in particular, $H=<a^r>$ so that $a=(a^r)^l$ for $l$ an integer). We cannot have $b^2in bH$ either because $b^2=ba^k$ yields $bin H$. Thus $b^2=1$.
Also $(ba)^2=1$. Indeed, if $babain bH$, then similarly $bin H$. If $babain H-1$, then $baba=a^rimplies bab=a^r-1implies b^2=a^r-2$, and as above, $a=(a^r-2)^l=(b^2)^limplies b=a^-2l$.
So $G$ is generated by $a,b$ and the relations $a^p=b^2=baba=1$ hold in $G$.
Is it sufficient to conclude that $G$ is isomorphic to $D_p$? Or do I need to verify somehow that there are no relations in $G$ other than those implied by the ones above? And are there any other gaps or mistakes in my proof?
abstract-algebra group-theory group-isomorphism group-presentation dihedral-groups
asked Aug 6 at 17:24
user437309
556212
asked Aug 6 at 17:24
user437309
556212
asked Aug 6 at 17:24
user437309
556212
asked Aug 6 at 17:24
user437309
556212
556212
1
How are you getting from $b^2l=a$ to $b=a^-2l$?
– Steve D
Aug 6 at 17:32
1
Here is one way to show $b^2neq a^r$: then $b$ would have order $2p$, and your group would be cyclic.
– Steve D
Aug 6 at 17:34
@SteveD Oops that's indeed wrong, thanks.
– user437309
Aug 6 at 17:36
add a comment |Â
1
How are you getting from $b^2l=a$ to $b=a^-2l$?
– Steve D
Aug 6 at 17:32
1
Here is one way to show $b^2neq a^r$: then $b$ would have order $2p$, and your group would be cyclic.
– Steve D
Aug 6 at 17:34
@SteveD Oops that's indeed wrong, thanks.
– user437309
Aug 6 at 17:36
1
1
How are you getting from $b^2l=a$ to $b=a^-2l$?
– Steve D
Aug 6 at 17:32
How are you getting from $b^2l=a$ to $b=a^-2l$?
– Steve D
Aug 6 at 17:32
1
1
Here is one way to show $b^2neq a^r$: then $b$ would have order $2p$, and your group would be cyclic.
– Steve D
Aug 6 at 17:34
Here is one way to show $b^2neq a^r$: then $b$ would have order $2p$, and your group would be cyclic.
– Steve D
Aug 6 at 17:34
@SteveD Oops that's indeed wrong, thanks.
– user437309
Aug 6 at 17:36
@SteveD Oops that's indeed wrong, thanks.
– user437309
Aug 6 at 17:36
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
You already know that every element has the form $a^ib^j$ for appropriate exponents $i,j$. Furthermore, you can calculate any product $a^ib^ja^i’b^j’$ using the conjugation rule $bab=a^-1$.
Therefore, you have the complete multiplication table for the group. Since the dihedral group has that same multiplication table, your unknown group is indeed isomorphic to the dihedral one.
answered Aug 6 at 17:33
Alon Amit
10.3k3765
I guess to say that every element is of the form $a^ib^j$ I have to prove that $ba^s=a^-sb$ (which follows easily from $ba=a^-1b$). Then every element $ba^i$ from $bH$ can be written as $a^-ib$; the square of any such element can be shown to be $1$; the product of an element from $H$ by an element from $bH$ viz. $a^iba^j$ can also be written as $a^ia^-jb=a^i-jb$, and $ba^ja^i=ba^i+j=a^-i-jb$. Finally, $a^ib^ja^kb^m$, according to my calculations is $a^ipm kb^m+j$ (using $ba^k=a^-kb$, not sure how to only use $bab=a^-1$).
– user437309
Aug 6 at 18:08
To say that every element has the form $a^ib^j$ you just need to count. There are $p$ elements $a^i$ and $p$ more of the form $a^ib$.
– Alon Amit
Aug 6 at 18:14
I’m not sure I follow the second question. Do you see how to compute the product of any two elements?
– Alon Amit
Aug 6 at 18:15
For your first comment: but we don't know a priori that elements from $HbH$ (those of the form $a^iba^j$) lie in $G$ (i.e., that $G$ is closed under multiplication), that's why I mentioned the argument with transforming them to the form $a^i-jb$. So I don't see why just counting works. (And as I'm writing this comment, I'm realizing that we also need to prove that the inverse of all products of elements of $G$ also have the form $a^ib^j$, this also seems a hassle...)
– user437309
Aug 6 at 18:23
For your second comment, I think I can see how to transform $a^ib^ja^kb^m$ to the form $a^rb^s$, but only if I use the relation $ba^q=a^-qb$, not just the relation $bab=a^-1$, which you mentioned.
– user437309
Aug 6 at 18:25
 |Â
show 7 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
You already know that every element has the form $a^ib^j$ for appropriate exponents $i,j$. Furthermore, you can calculate any product $a^ib^ja^i’b^j’$ using the conjugation rule $bab=a^-1$.
Therefore, you have the complete multiplication table for the group. Since the dihedral group has that same multiplication table, your unknown group is indeed isomorphic to the dihedral one.
answered Aug 6 at 17:33
Alon Amit
10.3k3765
I guess to say that every element is of the form $a^ib^j$ I have to prove that $ba^s=a^-sb$ (which follows easily from $ba=a^-1b$). Then every element $ba^i$ from $bH$ can be written as $a^-ib$; the square of any such element can be shown to be $1$; the product of an element from $H$ by an element from $bH$ viz. $a^iba^j$ can also be written as $a^ia^-jb=a^i-jb$, and $ba^ja^i=ba^i+j=a^-i-jb$. Finally, $a^ib^ja^kb^m$, according to my calculations is $a^ipm kb^m+j$ (using $ba^k=a^-kb$, not sure how to only use $bab=a^-1$).
– user437309
Aug 6 at 18:08
To say that every element has the form $a^ib^j$ you just need to count. There are $p$ elements $a^i$ and $p$ more of the form $a^ib$.
– Alon Amit
Aug 6 at 18:14
I’m not sure I follow the second question. Do you see how to compute the product of any two elements?
– Alon Amit
Aug 6 at 18:15
For your first comment: but we don't know a priori that elements from $HbH$ (those of the form $a^iba^j$) lie in $G$ (i.e., that $G$ is closed under multiplication), that's why I mentioned the argument with transforming them to the form $a^i-jb$. So I don't see why just counting works. (And as I'm writing this comment, I'm realizing that we also need to prove that the inverse of all products of elements of $G$ also have the form $a^ib^j$, this also seems a hassle...)
– user437309
Aug 6 at 18:23
For your second comment, I think I can see how to transform $a^ib^ja^kb^m$ to the form $a^rb^s$, but only if I use the relation $ba^q=a^-qb$, not just the relation $bab=a^-1$, which you mentioned.
– user437309
Aug 6 at 18:25
 |Â
show 7 more comments
up vote
2
down vote
accepted
You already know that every element has the form $a^ib^j$ for appropriate exponents $i,j$. Furthermore, you can calculate any product $a^ib^ja^i’b^j’$ using the conjugation rule $bab=a^-1$.
Therefore, you have the complete multiplication table for the group. Since the dihedral group has that same multiplication table, your unknown group is indeed isomorphic to the dihedral one.
answered Aug 6 at 17:33
Alon Amit
10.3k3765
I guess to say that every element is of the form $a^ib^j$ I have to prove that $ba^s=a^-sb$ (which follows easily from $ba=a^-1b$). Then every element $ba^i$ from $bH$ can be written as $a^-ib$; the square of any such element can be shown to be $1$; the product of an element from $H$ by an element from $bH$ viz. $a^iba^j$ can also be written as $a^ia^-jb=a^i-jb$, and $ba^ja^i=ba^i+j=a^-i-jb$. Finally, $a^ib^ja^kb^m$, according to my calculations is $a^ipm kb^m+j$ (using $ba^k=a^-kb$, not sure how to only use $bab=a^-1$).
– user437309
Aug 6 at 18:08
To say that every element has the form $a^ib^j$ you just need to count. There are $p$ elements $a^i$ and $p$ more of the form $a^ib$.
– Alon Amit
Aug 6 at 18:14
I’m not sure I follow the second question. Do you see how to compute the product of any two elements?
– Alon Amit
Aug 6 at 18:15
For your first comment: but we don't know a priori that elements from $HbH$ (those of the form $a^iba^j$) lie in $G$ (i.e., that $G$ is closed under multiplication), that's why I mentioned the argument with transforming them to the form $a^i-jb$. So I don't see why just counting works. (And as I'm writing this comment, I'm realizing that we also need to prove that the inverse of all products of elements of $G$ also have the form $a^ib^j$, this also seems a hassle...)
– user437309
Aug 6 at 18:23
For your second comment, I think I can see how to transform $a^ib^ja^kb^m$ to the form $a^rb^s$, but only if I use the relation $ba^q=a^-qb$, not just the relation $bab=a^-1$, which you mentioned.
– user437309
Aug 6 at 18:25
 |Â
show 7 more comments
up vote
2
down vote
accepted
up vote
2
down vote
accepted
You already know that every element has the form $a^ib^j$ for appropriate exponents $i,j$. Furthermore, you can calculate any product $a^ib^ja^i’b^j’$ using the conjugation rule $bab=a^-1$.
Therefore, you have the complete multiplication table for the group. Since the dihedral group has that same multiplication table, your unknown group is indeed isomorphic to the dihedral one.
answered Aug 6 at 17:33
Alon Amit
10.3k3765
You already know that every element has the form $a^ib^j$ for appropriate exponents $i,j$. Furthermore, you can calculate any product $a^ib^ja^i’b^j’$ using the conjugation rule $bab=a^-1$.
Therefore, you have the complete multiplication table for the group. Since the dihedral group has that same multiplication table, your unknown group is indeed isomorphic to the dihedral one.
answered Aug 6 at 17:33
Alon Amit
10.3k3765
answered Aug 6 at 17:33
Alon Amit
10.3k3765
answered Aug 6 at 17:33
Alon Amit
10.3k3765
answered Aug 6 at 17:33
Alon Amit
10.3k3765
10.3k3765
I guess to say that every element is of the form $a^ib^j$ I have to prove that $ba^s=a^-sb$ (which follows easily from $ba=a^-1b$). Then every element $ba^i$ from $bH$ can be written as $a^-ib$; the square of any such element can be shown to be $1$; the product of an element from $H$ by an element from $bH$ viz. $a^iba^j$ can also be written as $a^ia^-jb=a^i-jb$, and $ba^ja^i=ba^i+j=a^-i-jb$. Finally, $a^ib^ja^kb^m$, according to my calculations is $a^ipm kb^m+j$ (using $ba^k=a^-kb$, not sure how to only use $bab=a^-1$).
– user437309
Aug 6 at 18:08
To say that every element has the form $a^ib^j$ you just need to count. There are $p$ elements $a^i$ and $p$ more of the form $a^ib$.
– Alon Amit
Aug 6 at 18:14
I’m not sure I follow the second question. Do you see how to compute the product of any two elements?
– Alon Amit
Aug 6 at 18:15
For your first comment: but we don't know a priori that elements from $HbH$ (those of the form $a^iba^j$) lie in $G$ (i.e., that $G$ is closed under multiplication), that's why I mentioned the argument with transforming them to the form $a^i-jb$. So I don't see why just counting works. (And as I'm writing this comment, I'm realizing that we also need to prove that the inverse of all products of elements of $G$ also have the form $a^ib^j$, this also seems a hassle...)
– user437309
Aug 6 at 18:23
For your second comment, I think I can see how to transform $a^ib^ja^kb^m$ to the form $a^rb^s$, but only if I use the relation $ba^q=a^-qb$, not just the relation $bab=a^-1$, which you mentioned.
– user437309
Aug 6 at 18:25
 |Â
show 7 more comments
I guess to say that every element is of the form $a^ib^j$ I have to prove that $ba^s=a^-sb$ (which follows easily from $ba=a^-1b$). Then every element $ba^i$ from $bH$ can be written as $a^-ib$; the square of any such element can be shown to be $1$; the product of an element from $H$ by an element from $bH$ viz. $a^iba^j$ can also be written as $a^ia^-jb=a^i-jb$, and $ba^ja^i=ba^i+j=a^-i-jb$. Finally, $a^ib^ja^kb^m$, according to my calculations is $a^ipm kb^m+j$ (using $ba^k=a^-kb$, not sure how to only use $bab=a^-1$).
– user437309
Aug 6 at 18:08
To say that every element has the form $a^ib^j$ you just need to count. There are $p$ elements $a^i$ and $p$ more of the form $a^ib$.
– Alon Amit
Aug 6 at 18:14
I’m not sure I follow the second question. Do you see how to compute the product of any two elements?
– Alon Amit
Aug 6 at 18:15
For your first comment: but we don't know a priori that elements from $HbH$ (those of the form $a^iba^j$) lie in $G$ (i.e., that $G$ is closed under multiplication), that's why I mentioned the argument with transforming them to the form $a^i-jb$. So I don't see why just counting works. (And as I'm writing this comment, I'm realizing that we also need to prove that the inverse of all products of elements of $G$ also have the form $a^ib^j$, this also seems a hassle...)
– user437309
Aug 6 at 18:23
For your second comment, I think I can see how to transform $a^ib^ja^kb^m$ to the form $a^rb^s$, but only if I use the relation $ba^q=a^-qb$, not just the relation $bab=a^-1$, which you mentioned.
– user437309
Aug 6 at 18:25
I guess to say that every element is of the form $a^ib^j$ I have to prove that $ba^s=a^-sb$ (which follows easily from $ba=a^-1b$). Then every element $ba^i$ from $bH$ can be written as $a^-ib$; the square of any such element can be shown to be $1$; the product of an element from $H$ by an element from $bH$ viz. $a^iba^j$ can also be written as $a^ia^-jb=a^i-jb$, and $ba^ja^i=ba^i+j=a^-i-jb$. Finally, $a^ib^ja^kb^m$, according to my calculations is $a^ipm kb^m+j$ (using $ba^k=a^-kb$, not sure how to only use $bab=a^-1$).
– user437309
Aug 6 at 18:08
I guess to say that every element is of the form $a^ib^j$ I have to prove that $ba^s=a^-sb$ (which follows easily from $ba=a^-1b$). Then every element $ba^i$ from $bH$ can be written as $a^-ib$; the square of any such element can be shown to be $1$; the product of an element from $H$ by an element from $bH$ viz. $a^iba^j$ can also be written as $a^ia^-jb=a^i-jb$, and $ba^ja^i=ba^i+j=a^-i-jb$. Finally, $a^ib^ja^kb^m$, according to my calculations is $a^ipm kb^m+j$ (using $ba^k=a^-kb$, not sure how to only use $bab=a^-1$).
– user437309
Aug 6 at 18:08
To say that every element has the form $a^ib^j$ you just need to count. There are $p$ elements $a^i$ and $p$ more of the form $a^ib$.
– Alon Amit
Aug 6 at 18:14
To say that every element has the form $a^ib^j$ you just need to count. There are $p$ elements $a^i$ and $p$ more of the form $a^ib$.
– Alon Amit
Aug 6 at 18:14
I’m not sure I follow the second question. Do you see how to compute the product of any two elements?
– Alon Amit
Aug 6 at 18:15
I’m not sure I follow the second question. Do you see how to compute the product of any two elements?
– Alon Amit
Aug 6 at 18:15
For your first comment: but we don't know a priori that elements from $HbH$ (those of the form $a^iba^j$) lie in $G$ (i.e., that $G$ is closed under multiplication), that's why I mentioned the argument with transforming them to the form $a^i-jb$. So I don't see why just counting works. (And as I'm writing this comment, I'm realizing that we also need to prove that the inverse of all products of elements of $G$ also have the form $a^ib^j$, this also seems a hassle...)
– user437309
Aug 6 at 18:23
For your first comment: but we don't know a priori that elements from $HbH$ (those of the form $a^iba^j$) lie in $G$ (i.e., that $G$ is closed under multiplication), that's why I mentioned the argument with transforming them to the form $a^i-jb$. So I don't see why just counting works. (And as I'm writing this comment, I'm realizing that we also need to prove that the inverse of all products of elements of $G$ also have the form $a^ib^j$, this also seems a hassle...)
– user437309
Aug 6 at 18:23
For your second comment, I think I can see how to transform $a^ib^ja^kb^m$ to the form $a^rb^s$, but only if I use the relation $ba^q=a^-qb$, not just the relation $bab=a^-1$, which you mentioned.
– user437309
Aug 6 at 18:25
For your second comment, I think I can see how to transform $a^ib^ja^kb^m$ to the form $a^rb^s$, but only if I use the relation $ba^q=a^-qb$, not just the relation $bab=a^-1$, which you mentioned.
– user437309
Aug 6 at 18:25
 |Â
show 7 more comments
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2874120%2fgroups-of-order-2p%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
1
How are you getting from $b^2l=a$ to $b=a^-2l$?
– Steve D
Aug 6 at 17:32
1
Here is one way to show $b^2neq a^r$: then $b$ would have order $2p$, and your group would be cyclic.
– Steve D
Aug 6 at 17:34
@SteveD Oops that's indeed wrong, thanks.
– user437309
Aug 6 at 17:36