Mixing
project normal subgroup generated by a subgroup to its abelianization
$begingroup$
Say $operatorname{Ab}(G)$ is the abelianization of $G$. Let $G_1$ and $G_2$ be two groups, $G_1times G_2$ is the free product, then $G_1$ can be viewed as a subgroup in it. $j:G_1times G_2rightarrow operatorname{Ab}(G_1times G_2)$ is the natural homomorphism. $[G_1]$ is the normal subgroup generated by $G_1$ in $G_1times G_2$. It is claimed that $j([G_1])=j(G_1)$. Can anyone explain it more explicitly?
abstract-algebra normal-subgroups free-product
edited Jan 6 at 14:52
Thomas Shelby
2,525321
asked Jan 6 at 14:17
YijueYijue
31
$endgroup$
add a comment |
$begingroup$
Say $operatorname{Ab}(G)$ is the abelianization of $G$. Let $G_1$ and $G_2$ be two groups, $G_1times G_2$ is the free product, then $G_1$ can be viewed as a subgroup in it. $j:G_1times G_2rightarrow operatorname{Ab}(G_1times G_2)$ is the natural homomorphism. $[G_1]$ is the normal subgroup generated by $G_1$ in $G_1times G_2$. It is claimed that $j([G_1])=j(G_1)$. Can anyone explain it more explicitly?
abstract-algebra normal-subgroups free-product
edited Jan 6 at 14:52
Thomas Shelby
2,525321
asked Jan 6 at 14:17
YijueYijue
31
$endgroup$
add a comment |
$begingroup$
Say $operatorname{Ab}(G)$ is the abelianization of $G$. Let $G_1$ and $G_2$ be two groups, $G_1times G_2$ is the free product, then $G_1$ can be viewed as a subgroup in it. $j:G_1times G_2rightarrow operatorname{Ab}(G_1times G_2)$ is the natural homomorphism. $[G_1]$ is the normal subgroup generated by $G_1$ in $G_1times G_2$. It is claimed that $j([G_1])=j(G_1)$. Can anyone explain it more explicitly?
abstract-algebra normal-subgroups free-product
edited Jan 6 at 14:52
Thomas Shelby
2,525321
asked Jan 6 at 14:17
YijueYijue
31
$endgroup$
Say $operatorname{Ab}(G)$ is the abelianization of $G$. Let $G_1$ and $G_2$ be two groups, $G_1times G_2$ is the free product, then $G_1$ can be viewed as a subgroup in it. $j:G_1times G_2rightarrow operatorname{Ab}(G_1times G_2)$ is the natural homomorphism. $[G_1]$ is the normal subgroup generated by $G_1$ in $G_1times G_2$. It is claimed that $j([G_1])=j(G_1)$. Can anyone explain it more explicitly?
abstract-algebra normal-subgroups free-product
abstract-algebra normal-subgroups free-product
edited Jan 6 at 14:52
Thomas Shelby
2,525321
asked Jan 6 at 14:17
YijueYijue
31
edited Jan 6 at 14:52
Thomas Shelby
2,525321
asked Jan 6 at 14:17
YijueYijue
31
edited Jan 6 at 14:52
Thomas Shelby
2,525321
edited Jan 6 at 14:52
Thomas Shelby
2,525321
edited Jan 6 at 14:52
Thomas Shelby
2,525321
2,525321
asked Jan 6 at 14:17
YijueYijue
31
asked Jan 6 at 14:17
YijueYijue
31
asked Jan 6 at 14:17
YijueYijue
31
31
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Since it is not clear exactly which point of the question you don't understand, I'll try and over-explain and hopefully catch your sticking point.
The normal subgroup in $G$ generated by a subset $S$ of $G$ is the subgroup generated by the set of conjugators:
$$
left[ S right] = left< bigcup_{g in G} left{ g^{-1}sg mid s in S right} right>
$$
Then, using your notation, I hope it is clear that since $G_{1} subseteq left[ G_{1} right]$ we have $jleft(G_{1}right) subseteq jleft(left[ G_{1} right]right)$, so it remains to show the reverse inclusion. That is we need to show that
$$
jleft(left[ G_{1} right]right) subseteq jleft(G_{1}right)
$$
I hope it is clear that it suffices to show that
$$
jleft( g^{-1}hg right) in jleft(G_{1} right) text{for all} g in G_{1} * G_{2}, h in G_{1}.
$$
But given $g in G_{1} * G_{2}$, and $h in G_{1}$, since $j$ is a group homomorphism into an Abelian group we have
$$
jleft(g^{-1}hg right) = j(g)^{-1}j(h)j(g) = j(g)^{-1}j(g)j(h) = j(h) in jleft( G_{1}right).
$$
This concludes the proof. I hope this helps.
Just as a comment, notice that we haven't really used the specific structure of the free-product, or of the homomorphism $j$. In fact we have proven the following statement:
Let $phi : G rightarrow H$ be a group homomorphism with $H$ an Abelian group. Then for any subset $S subseteq H$, if $left[ S right]$ is the Normal subgroup in $G$ generated by $S$, then $phileft( left[S right] right) = phi(S)$.
answered Jan 6 at 14:43
Adam HigginsAdam Higgins
569112
$endgroup$
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
Since it is not clear exactly which point of the question you don't understand, I'll try and over-explain and hopefully catch your sticking point.
The normal subgroup in $G$ generated by a subset $S$ of $G$ is the subgroup generated by the set of conjugators:
$$
left[ S right] = left< bigcup_{g in G} left{ g^{-1}sg mid s in S right} right>
$$
Then, using your notation, I hope it is clear that since $G_{1} subseteq left[ G_{1} right]$ we have $jleft(G_{1}right) subseteq jleft(left[ G_{1} right]right)$, so it remains to show the reverse inclusion. That is we need to show that
$$
jleft(left[ G_{1} right]right) subseteq jleft(G_{1}right)
$$
I hope it is clear that it suffices to show that
$$
jleft( g^{-1}hg right) in jleft(G_{1} right) text{for all} g in G_{1} * G_{2}, h in G_{1}.
$$
But given $g in G_{1} * G_{2}$, and $h in G_{1}$, since $j$ is a group homomorphism into an Abelian group we have
$$
jleft(g^{-1}hg right) = j(g)^{-1}j(h)j(g) = j(g)^{-1}j(g)j(h) = j(h) in jleft( G_{1}right).
$$
This concludes the proof. I hope this helps.
Just as a comment, notice that we haven't really used the specific structure of the free-product, or of the homomorphism $j$. In fact we have proven the following statement:
Let $phi : G rightarrow H$ be a group homomorphism with $H$ an Abelian group. Then for any subset $S subseteq H$, if $left[ S right]$ is the Normal subgroup in $G$ generated by $S$, then $phileft( left[S right] right) = phi(S)$.
answered Jan 6 at 14:43
Adam HigginsAdam Higgins
569112
$endgroup$
add a comment |
$begingroup$
Since it is not clear exactly which point of the question you don't understand, I'll try and over-explain and hopefully catch your sticking point.
The normal subgroup in $G$ generated by a subset $S$ of $G$ is the subgroup generated by the set of conjugators:
$$
left[ S right] = left< bigcup_{g in G} left{ g^{-1}sg mid s in S right} right>
$$
Then, using your notation, I hope it is clear that since $G_{1} subseteq left[ G_{1} right]$ we have $jleft(G_{1}right) subseteq jleft(left[ G_{1} right]right)$, so it remains to show the reverse inclusion. That is we need to show that
$$
jleft(left[ G_{1} right]right) subseteq jleft(G_{1}right)
$$
I hope it is clear that it suffices to show that
$$
jleft( g^{-1}hg right) in jleft(G_{1} right) text{for all} g in G_{1} * G_{2}, h in G_{1}.
$$
But given $g in G_{1} * G_{2}$, and $h in G_{1}$, since $j$ is a group homomorphism into an Abelian group we have
$$
jleft(g^{-1}hg right) = j(g)^{-1}j(h)j(g) = j(g)^{-1}j(g)j(h) = j(h) in jleft( G_{1}right).
$$
This concludes the proof. I hope this helps.
Just as a comment, notice that we haven't really used the specific structure of the free-product, or of the homomorphism $j$. In fact we have proven the following statement:
Let $phi : G rightarrow H$ be a group homomorphism with $H$ an Abelian group. Then for any subset $S subseteq H$, if $left[ S right]$ is the Normal subgroup in $G$ generated by $S$, then $phileft( left[S right] right) = phi(S)$.
answered Jan 6 at 14:43
Adam HigginsAdam Higgins
569112
$endgroup$
add a comment |
$begingroup$
Since it is not clear exactly which point of the question you don't understand, I'll try and over-explain and hopefully catch your sticking point.
The normal subgroup in $G$ generated by a subset $S$ of $G$ is the subgroup generated by the set of conjugators:
$$
left[ S right] = left< bigcup_{g in G} left{ g^{-1}sg mid s in S right} right>
$$
Then, using your notation, I hope it is clear that since $G_{1} subseteq left[ G_{1} right]$ we have $jleft(G_{1}right) subseteq jleft(left[ G_{1} right]right)$, so it remains to show the reverse inclusion. That is we need to show that
$$
jleft(left[ G_{1} right]right) subseteq jleft(G_{1}right)
$$
I hope it is clear that it suffices to show that
$$
jleft( g^{-1}hg right) in jleft(G_{1} right) text{for all} g in G_{1} * G_{2}, h in G_{1}.
$$
But given $g in G_{1} * G_{2}$, and $h in G_{1}$, since $j$ is a group homomorphism into an Abelian group we have
$$
jleft(g^{-1}hg right) = j(g)^{-1}j(h)j(g) = j(g)^{-1}j(g)j(h) = j(h) in jleft( G_{1}right).
$$
This concludes the proof. I hope this helps.
Just as a comment, notice that we haven't really used the specific structure of the free-product, or of the homomorphism $j$. In fact we have proven the following statement:
Let $phi : G rightarrow H$ be a group homomorphism with $H$ an Abelian group. Then for any subset $S subseteq H$, if $left[ S right]$ is the Normal subgroup in $G$ generated by $S$, then $phileft( left[S right] right) = phi(S)$.
answered Jan 6 at 14:43
Adam HigginsAdam Higgins
569112
$endgroup$
Since it is not clear exactly which point of the question you don't understand, I'll try and over-explain and hopefully catch your sticking point.
The normal subgroup in $G$ generated by a subset $S$ of $G$ is the subgroup generated by the set of conjugators:
$$
left[ S right] = left< bigcup_{g in G} left{ g^{-1}sg mid s in S right} right>
$$
Then, using your notation, I hope it is clear that since $G_{1} subseteq left[ G_{1} right]$ we have $jleft(G_{1}right) subseteq jleft(left[ G_{1} right]right)$, so it remains to show the reverse inclusion. That is we need to show that
$$
jleft(left[ G_{1} right]right) subseteq jleft(G_{1}right)
$$
I hope it is clear that it suffices to show that
$$
jleft( g^{-1}hg right) in jleft(G_{1} right) text{for all} g in G_{1} * G_{2}, h in G_{1}.
$$
But given $g in G_{1} * G_{2}$, and $h in G_{1}$, since $j$ is a group homomorphism into an Abelian group we have
$$
jleft(g^{-1}hg right) = j(g)^{-1}j(h)j(g) = j(g)^{-1}j(g)j(h) = j(h) in jleft( G_{1}right).
$$
This concludes the proof. I hope this helps.
Just as a comment, notice that we haven't really used the specific structure of the free-product, or of the homomorphism $j$. In fact we have proven the following statement:
Let $phi : G rightarrow H$ be a group homomorphism with $H$ an Abelian group. Then for any subset $S subseteq H$, if $left[ S right]$ is the Normal subgroup in $G$ generated by $S$, then $phileft( left[S right] right) = phi(S)$.
answered Jan 6 at 14:43
Adam HigginsAdam Higgins
569112
answered Jan 6 at 14:43
Adam HigginsAdam Higgins
569112
answered Jan 6 at 14:43
Adam HigginsAdam Higgins
569112
answered Jan 6 at 14:43
Adam HigginsAdam Higgins
569112
569112
add a comment |
add a comment |
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