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Consider the following Group theoretical question:

Let $G$ be a group and $H \subset G$ a subgroup. Let $\rho : G \rightarrow H$ be a homomorphism such that the restriction of $\rho$ to $H$, $\rho|_H = \text{id}_H$. Let $a \in H$, $b \in \text{Ker}(\rho)$. Also suppose that $H \triangleleft G$, that is $H$ is normal in $G$. Proof that $ab = ba$.

I am rather stuck already at the beginning. Somehow I figure that the equivalent statement of $H \triangleleft G$, $\forall (g \in G, h \in H) : ghg^{-1} \in H$ is involved. Also I noticed that $\text{Ker}(\rho) \triangleleft G$ (a generally known result), $H \subset \text{Ker}(\rho)$, and therefore that in particular $H \triangleleft \text{Ker}(\rho)$. However after these simple facts I have no solution strategy, no clue how to "chop up" the problem so to speak. Arbitrarily substituting the equivalent condition for a group to be normal in another to hopefully magically obtain $ab = ba$ suddenly was also to no avail.

I must be missing something obvious. Any hints, please? Thanks a bunch in advance!

This is not homework...

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  • $\begingroup$ If you were at all curious, sometimes this homomorphism is referred to as an "inclusion map." $\endgroup$ Apr 9 '16 at 17:27
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Consider the element $bab^{-1}$, by normality of $H$ in $G$, this must be in $H$. Call it as $h$. Then $\rho(bab^{-1})=\rho(b)\rho(a)\rho(b^{-1})=\rho(h)$. But $\rho$ is identity on $H$, so this gives $a=h$.

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  • $\begingroup$ Aah, $\text{id}_H$ is the identity mapping, not the identity element of $H$. That explains a lot. Thanks! $\endgroup$
    – Jori
    Apr 9 '16 at 17:50
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Hint(s) : $$ab=ba \Leftrightarrow a=bab^{-1}$$ Both terms in this second equation are in $H$. Try proving that they have the same image under $\rho$.

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$H \ni a(ba^{-1}b^{-1}) = (aba^{-1})b^{-1} \in \ker \rho \implies aba^{-1}b^{-1} \in \ker \rho \cap H \implies aba^{-1}b^{-1} = e_{G}$

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