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Which of the following is correct a) $a^3= e$, $a^4 neq e$ b) $a^4 neq e$, $ a^5 neq e $
Clash Royale CLAN TAG#URR8PPP
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-1
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Let $G$ be group with identity element $e$ such that for some $a in G$, $a^2 neq e $ and $a^6 = e$
Then which of the following is correct?
a) $a^3= e$, $a^4 neq e$
b) $a^4 neq e$, $a^5 neq e $
I think option a) will be true because $a^6 = a^3.a^3 = e$ as $a^3= e$ and $a^4 neq e$
Is it correct?
Please help me
abstract-algebra
edited Jul 26 at 15:53
egreg
164k1180187
asked Jul 26 at 14:15
stupid
53518
 |Â
show 1 more comment
up vote
-1
down vote
favorite
Let $G$ be group with identity element $e$ such that for some $a in G$, $a^2 neq e $ and $a^6 = e$
Then which of the following is correct?
a) $a^3= e$, $a^4 neq e$
b) $a^4 neq e$, $a^5 neq e $
I think option a) will be true because $a^6 = a^3.a^3 = e$ as $a^3= e$ and $a^4 neq e$
Is it correct?
Please help me
abstract-algebra
edited Jul 26 at 15:53
egreg
164k1180187
asked Jul 26 at 14:15
stupid
53518
2
How can you justify the identity $a^6=a^3$? Why would it be true? (Actually, it is not true in general. There exists groups with elements of order exactly $6$. For instance, $1$ is of order $6$ in the additive group $(mathbbZ/6mathbbZ,+)$.
– Suzet
Jul 26 at 14:17
@ suzet as i take $a^6 = a^3.a^3 = 1.e$ where $a^3= 1 $ or $e$
– stupid
Jul 26 at 14:19
2
But you don't know whether $a^3=e$ or not, right? You only know $a^2 not =e$ and $a^6 = e$. You may then deduce that $(a^3)^2=e$, but not necessarily that $a^3=e$. Actually, the correct answer is $b)$. Can you see why?
– Suzet
Jul 26 at 14:21
im not getting @Suzet can u explain more
– stupid
Jul 26 at 14:26
2
I am going to write an answer to sum it up then.
– Suzet
Jul 26 at 14:26
 |Â
show 1 more comment
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Let $G$ be group with identity element $e$ such that for some $a in G$, $a^2 neq e $ and $a^6 = e$
Then which of the following is correct?
a) $a^3= e$, $a^4 neq e$
b) $a^4 neq e$, $a^5 neq e $
I think option a) will be true because $a^6 = a^3.a^3 = e$ as $a^3= e$ and $a^4 neq e$
Is it correct?
Please help me
abstract-algebra
edited Jul 26 at 15:53
egreg
164k1180187
asked Jul 26 at 14:15
stupid
53518
Let $G$ be group with identity element $e$ such that for some $a in G$, $a^2 neq e $ and $a^6 = e$
Then which of the following is correct?
a) $a^3= e$, $a^4 neq e$
b) $a^4 neq e$, $a^5 neq e $
I think option a) will be true because $a^6 = a^3.a^3 = e$ as $a^3= e$ and $a^4 neq e$
Is it correct?
Please help me
abstract-algebra
edited Jul 26 at 15:53
egreg
164k1180187
asked Jul 26 at 14:15
stupid
53518
edited Jul 26 at 15:53
egreg
164k1180187
edited Jul 26 at 15:53
egreg
164k1180187
edited Jul 26 at 15:53
egreg
164k1180187
164k1180187
asked Jul 26 at 14:15
stupid
53518
asked Jul 26 at 14:15
stupid
53518
asked Jul 26 at 14:15
stupid
53518
53518
2
How can you justify the identity $a^6=a^3$? Why would it be true? (Actually, it is not true in general. There exists groups with elements of order exactly $6$. For instance, $1$ is of order $6$ in the additive group $(mathbbZ/6mathbbZ,+)$.
– Suzet
Jul 26 at 14:17
@ suzet as i take $a^6 = a^3.a^3 = 1.e$ where $a^3= 1 $ or $e$
– stupid
Jul 26 at 14:19
2
But you don't know whether $a^3=e$ or not, right? You only know $a^2 not =e$ and $a^6 = e$. You may then deduce that $(a^3)^2=e$, but not necessarily that $a^3=e$. Actually, the correct answer is $b)$. Can you see why?
– Suzet
Jul 26 at 14:21
im not getting @Suzet can u explain more
– stupid
Jul 26 at 14:26
2
I am going to write an answer to sum it up then.
– Suzet
Jul 26 at 14:26
 |Â
show 1 more comment
2
How can you justify the identity $a^6=a^3$? Why would it be true? (Actually, it is not true in general. There exists groups with elements of order exactly $6$. For instance, $1$ is of order $6$ in the additive group $(mathbbZ/6mathbbZ,+)$.
– Suzet
Jul 26 at 14:17
@ suzet as i take $a^6 = a^3.a^3 = 1.e$ where $a^3= 1 $ or $e$
– stupid
Jul 26 at 14:19
2
But you don't know whether $a^3=e$ or not, right? You only know $a^2 not =e$ and $a^6 = e$. You may then deduce that $(a^3)^2=e$, but not necessarily that $a^3=e$. Actually, the correct answer is $b)$. Can you see why?
– Suzet
Jul 26 at 14:21
im not getting @Suzet can u explain more
– stupid
Jul 26 at 14:26
2
I am going to write an answer to sum it up then.
– Suzet
Jul 26 at 14:26
2
2
How can you justify the identity $a^6=a^3$? Why would it be true? (Actually, it is not true in general. There exists groups with elements of order exactly $6$. For instance, $1$ is of order $6$ in the additive group $(mathbbZ/6mathbbZ,+)$.
– Suzet
Jul 26 at 14:17
How can you justify the identity $a^6=a^3$? Why would it be true? (Actually, it is not true in general. There exists groups with elements of order exactly $6$. For instance, $1$ is of order $6$ in the additive group $(mathbbZ/6mathbbZ,+)$.
– Suzet
Jul 26 at 14:17
@ suzet as i take $a^6 = a^3.a^3 = 1.e$ where $a^3= 1 $ or $e$
– stupid
Jul 26 at 14:19
@ suzet as i take $a^6 = a^3.a^3 = 1.e$ where $a^3= 1 $ or $e$
– stupid
Jul 26 at 14:19
2
2
But you don't know whether $a^3=e$ or not, right? You only know $a^2 not =e$ and $a^6 = e$. You may then deduce that $(a^3)^2=e$, but not necessarily that $a^3=e$. Actually, the correct answer is $b)$. Can you see why?
– Suzet
Jul 26 at 14:21
But you don't know whether $a^3=e$ or not, right? You only know $a^2 not =e$ and $a^6 = e$. You may then deduce that $(a^3)^2=e$, but not necessarily that $a^3=e$. Actually, the correct answer is $b)$. Can you see why?
– Suzet
Jul 26 at 14:21
im not getting @Suzet can u explain more
– stupid
Jul 26 at 14:26
im not getting @Suzet can u explain more
– stupid
Jul 26 at 14:26
2
2
I am going to write an answer to sum it up then.
– Suzet
Jul 26 at 14:26
I am going to write an answer to sum it up then.
– Suzet
Jul 26 at 14:26
 |Â
show 1 more comment
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
First, let me explain why $b)$ is correct.
Generally speaking, given an element $g$ of finite order inside a group, we have, for any integer $kin mathbbZ$, the following equivalence
$$g^k=e Leftrightarrow operatornameord(g) textdivides ;k$$
where I denote by $operatornameord(g)$ the order of $g$.
So here, we know that $operatornameord(a)$ divides $6$, hence $operatornameord(a)$ is either $1$, $2$, $3$ or $6$.
However, it can not be $1$ nor $2$: it would be in contradiction with the fact that $a^2 not =e$.
So we know that $operatornameord(a)$ is either $3$ or $6$. Actually then, in each case, we may see that the statement $b)$ is true. If the order of $a$ is $6$, then of course $a^4$ and $a^5$ can not be equal to $e$ ; and if the order is $3$, then $a^4=a$ and $a^5=a^2$ can not be equal to $e$.
This is why $b)$ is the correct answer.
As for $a)$, in order to prove that in is not true in general, we need to provide a counter example. That is, we need to find a group together with an element $a$ such that $a^2 not =e$, $a^6 = e$, and ($a^3 not =e$ OR $a^4 =e$). What lies inside the parenthesis is just the negation of $a)$.
One may see that any element of order $6$ inside a group satisfies these conditions (the proposition inside the parenthesis being satisfied by $a^3 not = e$).
So the easiest example of a group with an element of order $6$ is, to my opinion, the additive group $(mathbbZ/6mathbbZ,+)$ with the element $1$. Indeed, we have $2times 1=2 not = 0$, we have $6times 1 = 6 =0$ and we have $3times 1 = 3 not = 0$ inside this group.
answered Jul 26 at 14:36
Suzet
2,203427
thanks u Suzets
– stupid
Jul 26 at 14:44
1
You're welcome :)
– Suzet
Jul 26 at 14:51
add a comment |Â
up vote
3
down vote
Since $a^6=e$, the order of $a$ can be $1$, $2$, $3$ or $6$.
However, since $a^2ne e$, we can reject the cases where the order is $1$ or $2$.
Suppose the order is $3$. Then $a^4=a^3a=ane e$. Moreover $a^5=a^3a^2=a^2ne e$. In this case both a) and b) hold.
Suppose the order is $6$. Then $e,a,a^2,a^3,a^4,a^5$ are all distinct, in particular, $a^3ne e$, $a^4ne e$ and $a^5ne e$. In this case a) doesn't hold, but b) holds.
Thus the information is insufficient to decide. In a cyclic group of order $3$, both a) and b) hold for a generator of the group.
answered Jul 26 at 16:01
egreg
164k1180187
add a comment |Â
up vote
2
down vote
$a^2a^4 = a^6 = e$. This means that $a^2$ is the inverse of $a^4$. Since $a^2$ is not the identity, $a^4$ isn't either.
From a similar argument, $aa^5 = e$, thus $a^5$ is not identity.
answered Jul 26 at 14:46
Kaynex
2,3261714
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
First, let me explain why $b)$ is correct.
Generally speaking, given an element $g$ of finite order inside a group, we have, for any integer $kin mathbbZ$, the following equivalence
$$g^k=e Leftrightarrow operatornameord(g) textdivides ;k$$
where I denote by $operatornameord(g)$ the order of $g$.
So here, we know that $operatornameord(a)$ divides $6$, hence $operatornameord(a)$ is either $1$, $2$, $3$ or $6$.
However, it can not be $1$ nor $2$: it would be in contradiction with the fact that $a^2 not =e$.
So we know that $operatornameord(a)$ is either $3$ or $6$. Actually then, in each case, we may see that the statement $b)$ is true. If the order of $a$ is $6$, then of course $a^4$ and $a^5$ can not be equal to $e$ ; and if the order is $3$, then $a^4=a$ and $a^5=a^2$ can not be equal to $e$.
This is why $b)$ is the correct answer.
As for $a)$, in order to prove that in is not true in general, we need to provide a counter example. That is, we need to find a group together with an element $a$ such that $a^2 not =e$, $a^6 = e$, and ($a^3 not =e$ OR $a^4 =e$). What lies inside the parenthesis is just the negation of $a)$.
One may see that any element of order $6$ inside a group satisfies these conditions (the proposition inside the parenthesis being satisfied by $a^3 not = e$).
So the easiest example of a group with an element of order $6$ is, to my opinion, the additive group $(mathbbZ/6mathbbZ,+)$ with the element $1$. Indeed, we have $2times 1=2 not = 0$, we have $6times 1 = 6 =0$ and we have $3times 1 = 3 not = 0$ inside this group.
answered Jul 26 at 14:36
Suzet
2,203427
thanks u Suzets
– stupid
Jul 26 at 14:44
1
You're welcome :)
– Suzet
Jul 26 at 14:51
add a comment |Â
up vote
2
down vote
accepted
First, let me explain why $b)$ is correct.
Generally speaking, given an element $g$ of finite order inside a group, we have, for any integer $kin mathbbZ$, the following equivalence
$$g^k=e Leftrightarrow operatornameord(g) textdivides ;k$$
where I denote by $operatornameord(g)$ the order of $g$.
So here, we know that $operatornameord(a)$ divides $6$, hence $operatornameord(a)$ is either $1$, $2$, $3$ or $6$.
However, it can not be $1$ nor $2$: it would be in contradiction with the fact that $a^2 not =e$.
So we know that $operatornameord(a)$ is either $3$ or $6$. Actually then, in each case, we may see that the statement $b)$ is true. If the order of $a$ is $6$, then of course $a^4$ and $a^5$ can not be equal to $e$ ; and if the order is $3$, then $a^4=a$ and $a^5=a^2$ can not be equal to $e$.
This is why $b)$ is the correct answer.
As for $a)$, in order to prove that in is not true in general, we need to provide a counter example. That is, we need to find a group together with an element $a$ such that $a^2 not =e$, $a^6 = e$, and ($a^3 not =e$ OR $a^4 =e$). What lies inside the parenthesis is just the negation of $a)$.
One may see that any element of order $6$ inside a group satisfies these conditions (the proposition inside the parenthesis being satisfied by $a^3 not = e$).
So the easiest example of a group with an element of order $6$ is, to my opinion, the additive group $(mathbbZ/6mathbbZ,+)$ with the element $1$. Indeed, we have $2times 1=2 not = 0$, we have $6times 1 = 6 =0$ and we have $3times 1 = 3 not = 0$ inside this group.
answered Jul 26 at 14:36
Suzet
2,203427
thanks u Suzets
– stupid
Jul 26 at 14:44
1
You're welcome :)
– Suzet
Jul 26 at 14:51
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
First, let me explain why $b)$ is correct.
Generally speaking, given an element $g$ of finite order inside a group, we have, for any integer $kin mathbbZ$, the following equivalence
$$g^k=e Leftrightarrow operatornameord(g) textdivides ;k$$
where I denote by $operatornameord(g)$ the order of $g$.
So here, we know that $operatornameord(a)$ divides $6$, hence $operatornameord(a)$ is either $1$, $2$, $3$ or $6$.
However, it can not be $1$ nor $2$: it would be in contradiction with the fact that $a^2 not =e$.
So we know that $operatornameord(a)$ is either $3$ or $6$. Actually then, in each case, we may see that the statement $b)$ is true. If the order of $a$ is $6$, then of course $a^4$ and $a^5$ can not be equal to $e$ ; and if the order is $3$, then $a^4=a$ and $a^5=a^2$ can not be equal to $e$.
This is why $b)$ is the correct answer.
As for $a)$, in order to prove that in is not true in general, we need to provide a counter example. That is, we need to find a group together with an element $a$ such that $a^2 not =e$, $a^6 = e$, and ($a^3 not =e$ OR $a^4 =e$). What lies inside the parenthesis is just the negation of $a)$.
One may see that any element of order $6$ inside a group satisfies these conditions (the proposition inside the parenthesis being satisfied by $a^3 not = e$).
So the easiest example of a group with an element of order $6$ is, to my opinion, the additive group $(mathbbZ/6mathbbZ,+)$ with the element $1$. Indeed, we have $2times 1=2 not = 0$, we have $6times 1 = 6 =0$ and we have $3times 1 = 3 not = 0$ inside this group.
answered Jul 26 at 14:36
Suzet
2,203427
First, let me explain why $b)$ is correct.
Generally speaking, given an element $g$ of finite order inside a group, we have, for any integer $kin mathbbZ$, the following equivalence
$$g^k=e Leftrightarrow operatornameord(g) textdivides ;k$$
where I denote by $operatornameord(g)$ the order of $g$.
So here, we know that $operatornameord(a)$ divides $6$, hence $operatornameord(a)$ is either $1$, $2$, $3$ or $6$.
However, it can not be $1$ nor $2$: it would be in contradiction with the fact that $a^2 not =e$.
So we know that $operatornameord(a)$ is either $3$ or $6$. Actually then, in each case, we may see that the statement $b)$ is true. If the order of $a$ is $6$, then of course $a^4$ and $a^5$ can not be equal to $e$ ; and if the order is $3$, then $a^4=a$ and $a^5=a^2$ can not be equal to $e$.
This is why $b)$ is the correct answer.
As for $a)$, in order to prove that in is not true in general, we need to provide a counter example. That is, we need to find a group together with an element $a$ such that $a^2 not =e$, $a^6 = e$, and ($a^3 not =e$ OR $a^4 =e$). What lies inside the parenthesis is just the negation of $a)$.
One may see that any element of order $6$ inside a group satisfies these conditions (the proposition inside the parenthesis being satisfied by $a^3 not = e$).
So the easiest example of a group with an element of order $6$ is, to my opinion, the additive group $(mathbbZ/6mathbbZ,+)$ with the element $1$. Indeed, we have $2times 1=2 not = 0$, we have $6times 1 = 6 =0$ and we have $3times 1 = 3 not = 0$ inside this group.
answered Jul 26 at 14:36
Suzet
2,203427
answered Jul 26 at 14:36
Suzet
2,203427
answered Jul 26 at 14:36
Suzet
2,203427
answered Jul 26 at 14:36
Suzet
2,203427
2,203427
thanks u Suzets
– stupid
Jul 26 at 14:44
1
You're welcome :)
– Suzet
Jul 26 at 14:51
add a comment |Â
thanks u Suzets
– stupid
Jul 26 at 14:44
1
You're welcome :)
– Suzet
Jul 26 at 14:51
thanks u Suzets
– stupid
Jul 26 at 14:44
thanks u Suzets
– stupid
Jul 26 at 14:44
1
1
You're welcome :)
– Suzet
Jul 26 at 14:51
You're welcome :)
– Suzet
Jul 26 at 14:51
add a comment |Â
up vote
3
down vote
Since $a^6=e$, the order of $a$ can be $1$, $2$, $3$ or $6$.
However, since $a^2ne e$, we can reject the cases where the order is $1$ or $2$.
Suppose the order is $3$. Then $a^4=a^3a=ane e$. Moreover $a^5=a^3a^2=a^2ne e$. In this case both a) and b) hold.
Suppose the order is $6$. Then $e,a,a^2,a^3,a^4,a^5$ are all distinct, in particular, $a^3ne e$, $a^4ne e$ and $a^5ne e$. In this case a) doesn't hold, but b) holds.
Thus the information is insufficient to decide. In a cyclic group of order $3$, both a) and b) hold for a generator of the group.
answered Jul 26 at 16:01
egreg
164k1180187
add a comment |Â
up vote
3
down vote
Since $a^6=e$, the order of $a$ can be $1$, $2$, $3$ or $6$.
However, since $a^2ne e$, we can reject the cases where the order is $1$ or $2$.
Suppose the order is $3$. Then $a^4=a^3a=ane e$. Moreover $a^5=a^3a^2=a^2ne e$. In this case both a) and b) hold.
Suppose the order is $6$. Then $e,a,a^2,a^3,a^4,a^5$ are all distinct, in particular, $a^3ne e$, $a^4ne e$ and $a^5ne e$. In this case a) doesn't hold, but b) holds.
Thus the information is insufficient to decide. In a cyclic group of order $3$, both a) and b) hold for a generator of the group.
answered Jul 26 at 16:01
egreg
164k1180187
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Since $a^6=e$, the order of $a$ can be $1$, $2$, $3$ or $6$.
However, since $a^2ne e$, we can reject the cases where the order is $1$ or $2$.
Suppose the order is $3$. Then $a^4=a^3a=ane e$. Moreover $a^5=a^3a^2=a^2ne e$. In this case both a) and b) hold.
Suppose the order is $6$. Then $e,a,a^2,a^3,a^4,a^5$ are all distinct, in particular, $a^3ne e$, $a^4ne e$ and $a^5ne e$. In this case a) doesn't hold, but b) holds.
Thus the information is insufficient to decide. In a cyclic group of order $3$, both a) and b) hold for a generator of the group.
answered Jul 26 at 16:01
egreg
164k1180187
Since $a^6=e$, the order of $a$ can be $1$, $2$, $3$ or $6$.
However, since $a^2ne e$, we can reject the cases where the order is $1$ or $2$.
Suppose the order is $3$. Then $a^4=a^3a=ane e$. Moreover $a^5=a^3a^2=a^2ne e$. In this case both a) and b) hold.
Suppose the order is $6$. Then $e,a,a^2,a^3,a^4,a^5$ are all distinct, in particular, $a^3ne e$, $a^4ne e$ and $a^5ne e$. In this case a) doesn't hold, but b) holds.
Thus the information is insufficient to decide. In a cyclic group of order $3$, both a) and b) hold for a generator of the group.
answered Jul 26 at 16:01
egreg
164k1180187
answered Jul 26 at 16:01
egreg
164k1180187
answered Jul 26 at 16:01
egreg
164k1180187
answered Jul 26 at 16:01
egreg
164k1180187
164k1180187
add a comment |Â
add a comment |Â
up vote
2
down vote
$a^2a^4 = a^6 = e$. This means that $a^2$ is the inverse of $a^4$. Since $a^2$ is not the identity, $a^4$ isn't either.
From a similar argument, $aa^5 = e$, thus $a^5$ is not identity.
answered Jul 26 at 14:46
Kaynex
2,3261714
add a comment |Â
up vote
2
down vote
$a^2a^4 = a^6 = e$. This means that $a^2$ is the inverse of $a^4$. Since $a^2$ is not the identity, $a^4$ isn't either.
From a similar argument, $aa^5 = e$, thus $a^5$ is not identity.
answered Jul 26 at 14:46
Kaynex
2,3261714
add a comment |Â
up vote
2
down vote
up vote
2
down vote
$a^2a^4 = a^6 = e$. This means that $a^2$ is the inverse of $a^4$. Since $a^2$ is not the identity, $a^4$ isn't either.
From a similar argument, $aa^5 = e$, thus $a^5$ is not identity.
answered Jul 26 at 14:46
Kaynex
2,3261714
$a^2a^4 = a^6 = e$. This means that $a^2$ is the inverse of $a^4$. Since $a^2$ is not the identity, $a^4$ isn't either.
From a similar argument, $aa^5 = e$, thus $a^5$ is not identity.
answered Jul 26 at 14:46
Kaynex
2,3261714
answered Jul 26 at 14:46
Kaynex
2,3261714
answered Jul 26 at 14:46
Kaynex
2,3261714
answered Jul 26 at 14:46
Kaynex
2,3261714
2,3261714
add a comment |Â
add a comment |Â
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2
How can you justify the identity $a^6=a^3$? Why would it be true? (Actually, it is not true in general. There exists groups with elements of order exactly $6$. For instance, $1$ is of order $6$ in the additive group $(mathbbZ/6mathbbZ,+)$.
– Suzet
Jul 26 at 14:17
@ suzet as i take $a^6 = a^3.a^3 = 1.e$ where $a^3= 1 $ or $e$
– stupid
Jul 26 at 14:19
2
But you don't know whether $a^3=e$ or not, right? You only know $a^2 not =e$ and $a^6 = e$. You may then deduce that $(a^3)^2=e$, but not necessarily that $a^3=e$. Actually, the correct answer is $b)$. Can you see why?
– Suzet
Jul 26 at 14:21
im not getting @Suzet can u explain more
– stupid
Jul 26 at 14:26
2
I am going to write an answer to sum it up then.
– Suzet
Jul 26 at 14:26