Proving homomorphism version of correspondence theorem

Question

Let $\alpha:G\to G_1$ be a group homomorphism and $K = \ker\,\alpha$.

1. if $H$ is a subgroup of $G$ that contains $K$ the $\alpha(H)$ is a subgroup of $\alpha(G)$.

I can't see why $H$ needs to contain $K $. In fact can't we find a group $H'$ wich contains $K$ and $\alpha(H) = \alpha(H')$ for any arbitrary subgroup $H$?

In the Factor Group version of theorem we need $K$ in $H$ to define $H/K$ but I don't see such a thing here .

Edit: I think $H'$ whould be $HK$ (as $K$ is normal in $G$ so $HK = KH$ so it is a subgroup and it contains $K$ and it's image is equal with image of $H$).

Answer 1

You are correct that the assumption that $H$ contains $K$ is not required in order to show that $\alpha(H)$ is a subgroup of $\alpha(G)$. The reason this is stated this way is to pave the way to the Correspondence Theorem (aka Lattice Isomorphism Theorem, aka Fourth Isomorphism Theorem) that states that there is a one to one inclusion preserving correspondence between subgroups of $\alpha(G)$ and subgroups of $G$ that contain $K$.

You are also correct that if $H$ is any subgroup of $G$, then $\alpha(H)=\alpha(HK)$. In a sense, this is part of the Second Isomorphism Theorem, that tells you that $\frac{H}{H\cap K}\cong \frac{HK}{K}$ for any subgroup $H$: note that $\frac{H}{H\cap K}$ is isomorphic to $\alpha(H)$ (by the First Isomorphism Theorem), and $\frac{HK}{K}$ is isomorphic to $\alpha(HK)$. A little bit of work shows that in fact the isomorphism asserted in the Second Isomorphism Theorem.

As I have written elsewhere, any homomorphism $\alpha\colon G\to G_1$ partitions the subgroups of $G$ into two classes:

1. The subgroups that contain $K=\ker(\alpha)$;
2. The subgroups that do not contain $K$.

The Lattice Isomorphism Theorem tells you what happens to the first class under $\alpha$, and how they are connected to the subgroups of $\alpha(G)$. The Second Isomorphism Theorem tells you that if $H$ is a subgroup in the second family, then the image of $H$ will behave in the same manner (in a natural way) as the image of $HK$ does.

So the reason that the theorem you are quoting restricts its attention to subgroups of $G$ that contain $K$ is because those subgroups will play the leading roles in the remaining three isomorphism theorems, and not because the assumption is somehow necessary for the conclusion to follow.