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Please help to understand this problem.

Let $G$ be a group, $H$ an abelian group, $\phi : G \rightarrow H$ a homomorphism. Show that $C(G) \lhd \mathrm{Ker}(\phi)$

I must be misunderstanding something, because what if we have the natural (bijective) homomorphism from $\mathbb{Z}$ to $\mathbb{Z}$. Then the kernel of the homomorphism is $0$ but the commutator subgroup of $\mathbb{Z}$ is $\mathbb{Z}$.

Any help would be appreciated.

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    $\begingroup$ What's the commutator subgroup of an abelian group? What are the elements of the form $ghg^{-1}h^{-1}$ in an abelian group? $\endgroup$ Feb 25, 2014 at 23:52

3 Answers 3

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It seems as if you are misunderstanding the definition of the commutator. It is not the center of the group.

Many times denoted $[G,G]$, it is defined to be the subgroup generated by the set $\{xyx^{-1}y^{-1} : x, y \in G\}$, where I am using multiplicative notation.

In your example $\mathbb{Z}$, an element in the commutator will be, for example with $x = 5$ and $y = 7$: $5 + 7 - 5 - 7$. You should be able to answer PVAL's question above, and this should give an indication of how to solve the problem.

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  • $\begingroup$ The subgroup generated by such expressions! $\endgroup$ Feb 26, 2014 at 1:07
  • $\begingroup$ That is hardly a subgroup! $\endgroup$
    – Pedro
    Feb 26, 2014 at 2:41
  • $\begingroup$ @TedShifrin Ack! Thank you! $\endgroup$
    – RghtHndSd
    Feb 26, 2014 at 3:09
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Claim If $x=[a,b]$ then $\varphi(x)=[\varphi(a),\varphi(b)]$.

Proof [insert proof here]

Corollary For every group morphism $\eta :G\to H$ with $H$ abelian, $[G,G]\subseteq \ker \eta$.

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  • $\begingroup$ This is related to a theorem that every normal subgroup has the commutator as a subset. In fact, because it is a theorem that N is a normal subgroup iff it is the kernel of a group homomorphism, it follows that one can deduce from your corollary that every normal subgroup must have the commutator as a subset. $\endgroup$
    – Fomalhaut
    Nov 19, 2016 at 4:06
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If $x\in C(G)$ then $x$ is of the form $x=aba^{-1}b^{-1}$ for some $a,b \in G$. Now $\phi (x)= \phi(a)\phi(b)\phi(a)^{-1}\phi(b)^{-1}$ is in $H$, and since $H$ is abelian we have $\phi(x)=\phi(a)\phi(a)^{-1}\phi(b)\phi(b)^{-1}=1$ and thus $x \in \ker(\phi)$.

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  • $\begingroup$ You perhaps meant $x\in \ker \phi$. $\endgroup$
    – Diogenes
    Feb 27, 2014 at 17:35
  • $\begingroup$ @Diogenes : of course. thanks! $\endgroup$
    – Ludolila
    Feb 27, 2014 at 18:23

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