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Let $H$ and $K$ be normal subgroups of a group $G$ with $H \subseteq K$. Define $\phi: G/H \rightarrow G/K$ by $\phi(Ha)=Ka$

Prove $\phi$ is a homomorphism.

We are given a function $\phi$, to prove $\phi$ is a homomorphism we need to show the property associated with a homomorphism which (if i am right) is:

$$\phi(HaHb)=\phi(Ha)\phi(Hb)$$

then

$$\phi(HaHb)=KaKb=\phi(Ha)\phi(Hb)$$

For some reason, it does not feel right. The only thing I could think of is the equation for the homomorphism is wrong but I'm not sure. Looking for a second opinion or any suggestions.

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    $\begingroup$ The reason it feels wrong is because it is.:) $\phi(HaHb)$ is not known to be $KaKb$ since your definition of $\phi$ only talks about $\phi(Hc)$ for one $c$. So what you want to consider is how to write $HaHb$ in such a way that you know what $\phi$ will send it to. Also one of the things you should prove (which you don't) is that $\phi$ is well defined. You have $Ha=Hb$ for different $a$ and $b$. You should show that choice of representative has no effect on the function. $\endgroup$
    – DRF
    Apr 20, 2015 at 5:04
  • $\begingroup$ We were asked to prove certain parts. The first one was to prove that $\phi$ is well defined and I already proved that. The second part is the homomorphism which is what I'm stuck at. Could it be $\phi(Hab)=Kab$ for $a,b$ in $G$? $\endgroup$
    – Mark
    Apr 20, 2015 at 5:10
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    $\begingroup$ Exactly. And then show that corresponds to what you want. $\endgroup$
    – DRF
    Apr 20, 2015 at 5:12
  • $\begingroup$ Let me just clarify, I need to show $\phi(Hab)=\phi(Ha)\phi(Hb)$? $\endgroup$
    – Mark
    Apr 20, 2015 at 5:19
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    $\begingroup$ Precisely. That's the definition of a homomorphism. So you need that $\phi(Hab)=KaKb$. $\endgroup$
    – DRF
    Apr 20, 2015 at 5:21

1 Answer 1

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We have

$$\begin{align} \phi(HaHb)&=\phi(Hab)\\ &=Kab\\ &=(Ka)(Kb)\\ &=\phi(Ha)\phi(Hb). \end{align}$$

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