Let $\varphi:G\to K$ be homomorphism, where $G$ is a finite group and $K$,arbitrary group. Suppose $H$ be a subgroup of $G$ Prove the following:

Question

Let $\varphi:G\to K$ be homomorphism, where $G$ is a finite group and $K$ is an arbitrary group. Suppose that $H$ be a subgroup of $G$. Prove that:

a) $[\varphi(G):\varphi(H)]$ divides $[G:H]$

b)$|\varphi(H)|$divides$|H|$

The only thing I understand about this problem is that a homorphism means that it preserved operation in its mapping, that if G is finite, there are a set number of elements,that if H is a subgroup of G, then it preserves all group qualities and is in H. I am also assuming that I must use the First Isomomorphism Theorem, which states If $f:G \to H$ is a homomorphism of groups, then $f$ induces an isomorphism $G/Kerf \cong Im f$ but I am not sure exactly what that is telling me to do.

Answer 1

Let $K$ denote the kernel of $\varphi$.

Then by the first isomorphism theorem, $\varphi(G) \cong G/K$, and so $|\varphi(G)| = |G|/|K|$.

Also, $H \cap K$ is the kernel of $\varphi$ restricted to $H$. So $\varphi(H) \cong H/(H \cap K)$, meaning $|\varphi(H)| = |H|/|H \cap K|$. This proves b). a) Follows from some algebra:

$$[\varphi(G):\varphi(H)] = \frac{|\varphi(G)|}{|\varphi(H)|} = \frac{|G|/|K|}{|H|/|H \cap K|} = \frac{|G|\cdot |H \cap K|}{|K| \cdot |H|} = \frac{|G|/|H|}{|K|/|H \cap K|} = \frac{[G:H]}{[K:H \cap K]}$$

Answer 2

For a) consider a disjoint union $G = x_1H \cup \dotsc \cup x_nH$. Now apply $\varphi$ on both sides. Note that the union is not necessarily disjoint anymore, but we still get the result.

For b) you can indeed use the isomorphism theorem, namely for $\varphi_{|H}$.