Show this is an isomorphism.

Question

I am trying to show that the mapping from $G \rightarrow G/H +G/K$, where + is the direct product is a homomorphism. The mapping is defined as $\phi(g)=(gH,gK)$. I have also showed that for the normal groups H and K, their intersection is also a normal subgroup of G. From that I can use that to show that $G/H\cap K$ is isomorphic to $Z_2+Z_2$. H and K are two distinct subgroups of G each with index 2.

Should I first show it is a homomorphism and then use to show it is isomorphic? Because their is a hint saying its tricky to show that $\phi$ is onto where $\phi$ is the homomorphism.

Answer 1

$\phi: G \longrightarrow G/H + G/K$ is a surjective homomorphism. By the first isomorphism theorem, $$G/ \ker \phi \cong G/H + G/K$$ Finally, note that $ \ker \phi = H \cap K$ , and you are done.

Edit: since the index of $H,K$ is 2, then $G/H \cong G/K \cong \mathbb{Z}/2 \mathbb{Z}$. So $$G/H + G/K \cong \mathbb{Z}/2 \mathbb{Z} +\mathbb{Z}/2 \mathbb{Z}$$

Answer 2

I'm assuming you have trouble to show that the map $\phi$ is surjective. Here is a way to do it:

Since $H, K$ are two distinct subgroups, there is $k \in K -H.$ Since $G/H \cong \mathbb{Z}_2,$ we can assume that it is generated by $kH.$ Since $H$ is normal in $G$, $HK$ is a subgroup of $G$. Let $g \in G$. Then $gH = kH \Rightarrow g = kh,$ for some $h \in H$. Thus $G = HK.$ This shows that the map $\phi$ is surjective.