Homomorphism and indexes of subgroups

Question

Let $G$ be a finite group, $f:G \to M$ group homomorphism and $H \leq G$. Show that $[f(G):f(H)]$ divides $[G:H]$.

I have already shown that $f(H) \leq f(G) \leq M$. I guess I should use Lagrange's theorem somehow, but I don't have any other ideas on how to proceed. Any help would be appreciated!

Answer 1

With $K=\ker f$ and $L=H\cap K$ (which is a subgroup of $K$)

$$G/K\cong f(G)$$ $$H/L\cong f(H)$$

Thus with $|G|=g$ etcetera we get that $h/l$ divides $g/k$ and the index $$[f(G):f(H)]=\dfrac{g/k}{h/l}$$ divides $[G:H]=g/h$. Indeed, with $k=lm$ we get that it is $[G:H]/{m}$.

Answer 2

Sketch: Let $K$ be the kernel of $f$.

1. Since $K$ is normal, $HK$ is a subgroup of $G$.

2. There is a bijection between cosets of $HK$ in $G$ and cosets of $f(H)$ in $f(G)$. The bijection is given by $gHK$ corresponds to $f(g)f(H)$. To see where this comes from, observe that $f^{-1}(f(H))=HK$.

One direction of the proof: Suppose that the cosets $f(g_1)f(H)$ and $f(g_2)f(H)$ are equal. Then, there exists some $h\in H$ such that $f(g_1)f(h)=f(g_2)$. Therefore, $f(g_2^{-1}g_1h)=e_M$. Therefore, $g_2^{-1}g_1h\in K$ so there is some $k\in K$ such that $g_2^{-1}g_1h=k$ or that $g_2^{-1}g_1=kh^{-1}\in KH$. This implies that $g_2HK=g_1HK$.

Starting with $g_2^{-1}g_1\in HK$, write $g_1=g_2hk$ and use this to show that $f(g_1)f(H)=f(g_2)f(H)$ by a similar argument.

3. The result follows since $[G:H]=[G:HK][HK:H]$ and $[G:HK]=[f(G):f(H)]$.

Answer 3

let $[G:H]=n$ and $$ G = H + Hg_1 + \dots +Hg_n $$ so if we write $\bar x$ for $f(x)$ $$ \bar G = \bar H + \bar H \bar g_1 + \dots +\bar H\bar g_n $$ now $\bar H \bar g_j = \bar H \bar g_k$ iff $\bar g_j \bar g_k^{-1} \in \bar H$ iff $g_j g_k^{-1} \in H \cup K$ so $$ [\bar G:\bar H] = [G:H\cup K] $$ but we have: $$ [G:H] = [G:H \cup K][H \cup K:H] $$ so $$ [G:H] = [\bar G: \bar H][H \cup K:H] $$