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Subgroup generated by subset of group
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I had come across following term 'subgroup generated by subset' in Herstein Excercise.
Given any group G and subset U ,$barU$ be smallest subgroup containg that.
I thought that can be given by $barU$=$cap_Uin G_iG_i$ where $G_i$ is subgroup in G .Is there in other defination exist for $barU$.
Next question says to prove that $barU$ is normal if $gug^-1in U $ $forall gin G$
To prove that this I have to show that $gbarUg^-1=barU$
By defination above $gcap_Uin G_iG_i g^-1$=$cap_Uin G_igG_ig^-1$=$cap_Uin G_iG_i$
I had not concrete argument to write last thing .$G_i$ is subgroup containg U.
Please help me to find correct argument to write that .
abstract-algebra group-theory normal-subgroups
asked 2 days ago
SRJ
984216
add a comment |Â
up vote
1
down vote
favorite
I had come across following term 'subgroup generated by subset' in Herstein Excercise.
Given any group G and subset U ,$barU$ be smallest subgroup containg that.
I thought that can be given by $barU$=$cap_Uin G_iG_i$ where $G_i$ is subgroup in G .Is there in other defination exist for $barU$.
Next question says to prove that $barU$ is normal if $gug^-1in U $ $forall gin G$
To prove that this I have to show that $gbarUg^-1=barU$
By defination above $gcap_Uin G_iG_i g^-1$=$cap_Uin G_igG_ig^-1$=$cap_Uin G_iG_i$
I had not concrete argument to write last thing .$G_i$ is subgroup containg U.
Please help me to find correct argument to write that .
abstract-algebra group-theory normal-subgroups
asked 2 days ago
SRJ
984216
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I had come across following term 'subgroup generated by subset' in Herstein Excercise.
Given any group G and subset U ,$barU$ be smallest subgroup containg that.
I thought that can be given by $barU$=$cap_Uin G_iG_i$ where $G_i$ is subgroup in G .Is there in other defination exist for $barU$.
Next question says to prove that $barU$ is normal if $gug^-1in U $ $forall gin G$
To prove that this I have to show that $gbarUg^-1=barU$
By defination above $gcap_Uin G_iG_i g^-1$=$cap_Uin G_igG_ig^-1$=$cap_Uin G_iG_i$
I had not concrete argument to write last thing .$G_i$ is subgroup containg U.
Please help me to find correct argument to write that .
abstract-algebra group-theory normal-subgroups
asked 2 days ago
SRJ
984216
I had come across following term 'subgroup generated by subset' in Herstein Excercise.
Given any group G and subset U ,$barU$ be smallest subgroup containg that.
I thought that can be given by $barU$=$cap_Uin G_iG_i$ where $G_i$ is subgroup in G .Is there in other defination exist for $barU$.
Next question says to prove that $barU$ is normal if $gug^-1in U $ $forall gin G$
To prove that this I have to show that $gbarUg^-1=barU$
By defination above $gcap_Uin G_iG_i g^-1$=$cap_Uin G_igG_ig^-1$=$cap_Uin G_iG_i$
I had not concrete argument to write last thing .$G_i$ is subgroup containg U.
Please help me to find correct argument to write that .
abstract-algebra group-theory normal-subgroups
asked 2 days ago
SRJ
984216
asked 2 days ago
SRJ
984216
asked 2 days ago
SRJ
984216
asked 2 days ago
SRJ
984216
984216
add a comment |Â
add a comment |Â
1 Answer
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up vote
1
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The definition $barU = cap_U in G_i G_i$ is the standard definition for the subgroup generated by the subset $U$. Another way to define it is to construct it explicitly. You can definite $barU$ as the subset of all finite products of the type $u_1^k_1u_2^k_2cdots u_r^k_r$, where $u_i in U, k_i in mathbbZ$. However you have to show that this is a subgroup, which shouldn't be hard using the standard method.
You can use the explicit construction for the last part. For any $baru in barU$ we have that $baru = u_1^k_1u_2^k_2cdots u_r^k_r$ for some. $u_i in U, k_i in mathbbZ$. Then we have:
$$gbarug^-1 = gu_1^k_1u_2^k_2cdots u_r^k_rg^-1 = (gu_1g^-1)^k_1(gu_2g^-1)^k_2cdots (gu_rg^-1)^k_r = v_1^k_1v_2^k_2cdots v_r^k_r in barU$$
Hence the proof that $barU$ is normal if $gug^-1 in U$ for every $g in G, u in U$
answered 2 days ago
Stefan4024
27.7k52974
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The definition $barU = cap_U in G_i G_i$ is the standard definition for the subgroup generated by the subset $U$. Another way to define it is to construct it explicitly. You can definite $barU$ as the subset of all finite products of the type $u_1^k_1u_2^k_2cdots u_r^k_r$, where $u_i in U, k_i in mathbbZ$. However you have to show that this is a subgroup, which shouldn't be hard using the standard method.
You can use the explicit construction for the last part. For any $baru in barU$ we have that $baru = u_1^k_1u_2^k_2cdots u_r^k_r$ for some. $u_i in U, k_i in mathbbZ$. Then we have:
$$gbarug^-1 = gu_1^k_1u_2^k_2cdots u_r^k_rg^-1 = (gu_1g^-1)^k_1(gu_2g^-1)^k_2cdots (gu_rg^-1)^k_r = v_1^k_1v_2^k_2cdots v_r^k_r in barU$$
Hence the proof that $barU$ is normal if $gug^-1 in U$ for every $g in G, u in U$
answered 2 days ago
Stefan4024
27.7k52974
add a comment |Â
up vote
1
down vote
accepted
The definition $barU = cap_U in G_i G_i$ is the standard definition for the subgroup generated by the subset $U$. Another way to define it is to construct it explicitly. You can definite $barU$ as the subset of all finite products of the type $u_1^k_1u_2^k_2cdots u_r^k_r$, where $u_i in U, k_i in mathbbZ$. However you have to show that this is a subgroup, which shouldn't be hard using the standard method.
You can use the explicit construction for the last part. For any $baru in barU$ we have that $baru = u_1^k_1u_2^k_2cdots u_r^k_r$ for some. $u_i in U, k_i in mathbbZ$. Then we have:
$$gbarug^-1 = gu_1^k_1u_2^k_2cdots u_r^k_rg^-1 = (gu_1g^-1)^k_1(gu_2g^-1)^k_2cdots (gu_rg^-1)^k_r = v_1^k_1v_2^k_2cdots v_r^k_r in barU$$
Hence the proof that $barU$ is normal if $gug^-1 in U$ for every $g in G, u in U$
answered 2 days ago
Stefan4024
27.7k52974
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The definition $barU = cap_U in G_i G_i$ is the standard definition for the subgroup generated by the subset $U$. Another way to define it is to construct it explicitly. You can definite $barU$ as the subset of all finite products of the type $u_1^k_1u_2^k_2cdots u_r^k_r$, where $u_i in U, k_i in mathbbZ$. However you have to show that this is a subgroup, which shouldn't be hard using the standard method.
You can use the explicit construction for the last part. For any $baru in barU$ we have that $baru = u_1^k_1u_2^k_2cdots u_r^k_r$ for some. $u_i in U, k_i in mathbbZ$. Then we have:
$$gbarug^-1 = gu_1^k_1u_2^k_2cdots u_r^k_rg^-1 = (gu_1g^-1)^k_1(gu_2g^-1)^k_2cdots (gu_rg^-1)^k_r = v_1^k_1v_2^k_2cdots v_r^k_r in barU$$
Hence the proof that $barU$ is normal if $gug^-1 in U$ for every $g in G, u in U$
answered 2 days ago
Stefan4024
27.7k52974
The definition $barU = cap_U in G_i G_i$ is the standard definition for the subgroup generated by the subset $U$. Another way to define it is to construct it explicitly. You can definite $barU$ as the subset of all finite products of the type $u_1^k_1u_2^k_2cdots u_r^k_r$, where $u_i in U, k_i in mathbbZ$. However you have to show that this is a subgroup, which shouldn't be hard using the standard method.
You can use the explicit construction for the last part. For any $baru in barU$ we have that $baru = u_1^k_1u_2^k_2cdots u_r^k_r$ for some. $u_i in U, k_i in mathbbZ$. Then we have:
$$gbarug^-1 = gu_1^k_1u_2^k_2cdots u_r^k_rg^-1 = (gu_1g^-1)^k_1(gu_2g^-1)^k_2cdots (gu_rg^-1)^k_r = v_1^k_1v_2^k_2cdots v_r^k_r in barU$$
Hence the proof that $barU$ is normal if $gug^-1 in U$ for every $g in G, u in U$
answered 2 days ago
Stefan4024
27.7k52974
answered 2 days ago
Stefan4024
27.7k52974
answered 2 days ago
Stefan4024
27.7k52974
answered 2 days ago
Stefan4024
27.7k52974
27.7k52974
add a comment |Â
add a comment |Â
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