$Z(G)=left langle e right rangle$ if and only if $N_Gamma...
Let $G$ a group and let a group $Gamma=Gtimes G$ with $forall (a,b),(c,d)inGamma, (a,b)(c,d):=(ac,bd)$. Let a map $varphi:GtoGamma$ by $varphi(g)=(g,g)$. Prove that $Z(G)=left langle e right rangle$ if and only if $N_Gamma (Im(varphi))=Im(varphi)$.
Define $Q:=operatorname{Im}(varphi)$. I've shown that $Q={(g,g):gin G}$, $varphi$ is a homomorphism and that $G$ is an abelian group if and only if $Q$ is a normal subgroup of $Gamma$.
abstract-algebra group-theory group-homomorphism
edited Nov 28 at 19:36
Bernard
117k637111
asked Nov 28 at 17:32
J. Doe
956
closed as off-topic by Derek Holt, amWhy, Leucippus, KReiser, max_zorn Nov 29 at 6:24
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Let $G$ a group and let a group $Gamma=Gtimes G$ with $forall (a,b),(c,d)inGamma, (a,b)(c,d):=(ac,bd)$. Let a map $varphi:GtoGamma$ by $varphi(g)=(g,g)$. Prove that $Z(G)=left langle e right rangle$ if and only if $N_Gamma (Im(varphi))=Im(varphi)$.
Define $Q:=operatorname{Im}(varphi)$. I've shown that $Q={(g,g):gin G}$, $varphi$ is a homomorphism and that $G$ is an abelian group if and only if $Q$ is a normal subgroup of $Gamma$.
abstract-algebra group-theory group-homomorphism
edited Nov 28 at 19:36
Bernard
117k637111
asked Nov 28 at 17:32
J. Doe
956
closed as off-topic by Derek Holt, amWhy, Leucippus, KReiser, max_zorn Nov 29 at 6:24
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Derek Holt, amWhy, Leucippus, KReiser, max_zorn
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
Let $G$ a group and let a group $Gamma=Gtimes G$ with $forall (a,b),(c,d)inGamma, (a,b)(c,d):=(ac,bd)$. Let a map $varphi:GtoGamma$ by $varphi(g)=(g,g)$. Prove that $Z(G)=left langle e right rangle$ if and only if $N_Gamma (Im(varphi))=Im(varphi)$.
Define $Q:=operatorname{Im}(varphi)$. I've shown that $Q={(g,g):gin G}$, $varphi$ is a homomorphism and that $G$ is an abelian group if and only if $Q$ is a normal subgroup of $Gamma$.
abstract-algebra group-theory group-homomorphism
edited Nov 28 at 19:36
Bernard
117k637111
asked Nov 28 at 17:32
J. Doe
956
Let $G$ a group and let a group $Gamma=Gtimes G$ with $forall (a,b),(c,d)inGamma, (a,b)(c,d):=(ac,bd)$. Let a map $varphi:GtoGamma$ by $varphi(g)=(g,g)$. Prove that $Z(G)=left langle e right rangle$ if and only if $N_Gamma (Im(varphi))=Im(varphi)$.
Define $Q:=operatorname{Im}(varphi)$. I've shown that $Q={(g,g):gin G}$, $varphi$ is a homomorphism and that $G$ is an abelian group if and only if $Q$ is a normal subgroup of $Gamma$.
abstract-algebra group-theory group-homomorphism
abstract-algebra group-theory group-homomorphism
edited Nov 28 at 19:36
Bernard
117k637111
asked Nov 28 at 17:32
J. Doe
956
edited Nov 28 at 19:36
Bernard
117k637111
asked Nov 28 at 17:32
J. Doe
956
edited Nov 28 at 19:36
Bernard
117k637111
edited Nov 28 at 19:36
Bernard
117k637111
edited Nov 28 at 19:36
Bernard
117k637111
117k637111
asked Nov 28 at 17:32
J. Doe
956
asked Nov 28 at 17:32
J. Doe
956
asked Nov 28 at 17:32
J. Doe
956
956
closed as off-topic by Derek Holt, amWhy, Leucippus, KReiser, max_zorn Nov 29 at 6:24
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Derek Holt, amWhy, Leucippus, KReiser, max_zorn
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Derek Holt, amWhy, Leucippus, KReiser, max_zorn Nov 29 at 6:24
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Derek Holt, amWhy, Leucippus, KReiser, max_zorn
If this question can be reworded to fit the rules in the help center, please edit the question.
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1 Answer
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Suppose $Z(G)=langle erangle$ and let $(a,b)in N_Gamma(operatorname{Im}(phi))$. Then for all $gin G$ we have
$$operatorname{Im}(phi)ni (a,b)(g,g)(a,b)^{-1}=(aga^{-1},bgb^{-1}),$$
which (by your calculation of $operatorname{Im}(phi)$) means that for some $g'in G$ we have
$$(aga^{-1},bgb^{-1})=(g',g')implies aga^{-1}=bgb^{-1}.$$
Rearrange this equation to see that $a^{-1}b$ commutes with $g$, and since this holds for every $gin G$ we have $a^{-1}bin Z(G)=langle erangle$, which implies $a=b$ so $(a,b)in operatorname{Im}(phi)$.
edited Nov 28 at 19:37
Bernard
117k637111
answered Nov 28 at 17:44
Alex Mathers
10.7k21344
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Suppose $Z(G)=langle erangle$ and let $(a,b)in N_Gamma(operatorname{Im}(phi))$. Then for all $gin G$ we have
$$operatorname{Im}(phi)ni (a,b)(g,g)(a,b)^{-1}=(aga^{-1},bgb^{-1}),$$
which (by your calculation of $operatorname{Im}(phi)$) means that for some $g'in G$ we have
$$(aga^{-1},bgb^{-1})=(g',g')implies aga^{-1}=bgb^{-1}.$$
Rearrange this equation to see that $a^{-1}b$ commutes with $g$, and since this holds for every $gin G$ we have $a^{-1}bin Z(G)=langle erangle$, which implies $a=b$ so $(a,b)in operatorname{Im}(phi)$.
edited Nov 28 at 19:37
Bernard
117k637111
answered Nov 28 at 17:44
Alex Mathers
10.7k21344
add a comment |
Suppose $Z(G)=langle erangle$ and let $(a,b)in N_Gamma(operatorname{Im}(phi))$. Then for all $gin G$ we have
$$operatorname{Im}(phi)ni (a,b)(g,g)(a,b)^{-1}=(aga^{-1},bgb^{-1}),$$
which (by your calculation of $operatorname{Im}(phi)$) means that for some $g'in G$ we have
$$(aga^{-1},bgb^{-1})=(g',g')implies aga^{-1}=bgb^{-1}.$$
Rearrange this equation to see that $a^{-1}b$ commutes with $g$, and since this holds for every $gin G$ we have $a^{-1}bin Z(G)=langle erangle$, which implies $a=b$ so $(a,b)in operatorname{Im}(phi)$.
edited Nov 28 at 19:37
Bernard
117k637111
answered Nov 28 at 17:44
Alex Mathers
10.7k21344
add a comment |
Suppose $Z(G)=langle erangle$ and let $(a,b)in N_Gamma(operatorname{Im}(phi))$. Then for all $gin G$ we have
$$operatorname{Im}(phi)ni (a,b)(g,g)(a,b)^{-1}=(aga^{-1},bgb^{-1}),$$
which (by your calculation of $operatorname{Im}(phi)$) means that for some $g'in G$ we have
$$(aga^{-1},bgb^{-1})=(g',g')implies aga^{-1}=bgb^{-1}.$$
Rearrange this equation to see that $a^{-1}b$ commutes with $g$, and since this holds for every $gin G$ we have $a^{-1}bin Z(G)=langle erangle$, which implies $a=b$ so $(a,b)in operatorname{Im}(phi)$.
edited Nov 28 at 19:37
Bernard
117k637111
answered Nov 28 at 17:44
Alex Mathers
10.7k21344
Suppose $Z(G)=langle erangle$ and let $(a,b)in N_Gamma(operatorname{Im}(phi))$. Then for all $gin G$ we have
$$operatorname{Im}(phi)ni (a,b)(g,g)(a,b)^{-1}=(aga^{-1},bgb^{-1}),$$
which (by your calculation of $operatorname{Im}(phi)$) means that for some $g'in G$ we have
$$(aga^{-1},bgb^{-1})=(g',g')implies aga^{-1}=bgb^{-1}.$$
Rearrange this equation to see that $a^{-1}b$ commutes with $g$, and since this holds for every $gin G$ we have $a^{-1}bin Z(G)=langle erangle$, which implies $a=b$ so $(a,b)in operatorname{Im}(phi)$.
edited Nov 28 at 19:37
Bernard
117k637111
answered Nov 28 at 17:44
Alex Mathers
10.7k21344
edited Nov 28 at 19:37
Bernard
117k637111
edited Nov 28 at 19:37
Bernard
117k637111
edited Nov 28 at 19:37
Bernard
117k637111
117k637111
answered Nov 28 at 17:44
Alex Mathers
10.7k21344
answered Nov 28 at 17:44
Alex Mathers
10.7k21344
answered Nov 28 at 17:44
Alex Mathers
10.7k21344
10.7k21344
add a comment |
add a comment |
