# Prove $\phi$ is a homomorphism.

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.

• 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.
– DRF
Apr 20, 2015 at 5:04
• 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 $G$?
– Mark
Apr 20, 2015 at 5:10
• Exactly. And then show that corresponds to what you want.
– DRF
Apr 20, 2015 at 5:12
• Let me just clarify, I need to show $\phi(Hab)=\phi(Ha)\phi(Hb)$?
– Mark
Apr 20, 2015 at 5:19
• Precisely. That's the definition of a homomorphism. So you need that $\phi(Hab)=KaKb$.
– DRF
Apr 20, 2015 at 5:21

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