Kernel of composition of homomorphism Given $\theta$ is a homormophism from $G$ to $H$, and $\sigma$ is a homormophism from $H$ to $K$. How is Ker($\sigma\theta$) is related to Ker($\theta$)?If $\theta$ and $\sigma$ are onto and $G$ is finite, compute [Ker($\sigma\theta$):Ker($\theta$)] in terms of $|H|$ and $|K|$.
My progress: Let Ker($\theta$)$=\left\{g\in G\ |\ \theta(g)=e_H\right\}$ and Ker($\sigma\theta$)$=\left\{h\in G\ |\ \sigma\theta(h)=e_K\right\}$. Since $\sigma\theta$ is a homomorphism from $G$ to $K$, $\sigma\theta(h)= \sigma(\theta(h))=e_K=\sigma(\theta(g))$ where $g\in$ Ker($\theta$). Thus $\theta(h)$Ker($\sigma$)$=\theta(g)$Ker($\sigma$), so $\theta(h)$Ker($\sigma$)$=$Ker($\sigma$), so $h\in Ker(\theta)$ (since $\theta(h)=e_H$). Thus Ker($\sigma\theta$) $\leq$ Ker($\theta$). But this is clearly not true based on the definition of the two kernels! 
I'm pretty sure the other way around is true, which means Ker($\theta$) is a subgroup of Ker($\sigma\theta$), but I don't see why my proof above is incorrect. 
For the last part of the question, I think the result should be $|K|$ divides by $|H|$ (is this true though?), if we can show Ker($\theta$) is a subgroup of Ker($\sigma\theta$), which I haven't been able to prove yet:(
 A: 
*

*First of all:
$$g\in\ker(\theta)\ \Longrightarrow\ \theta(g)=e_{H}\ \Longrightarrow\ (\sigma\theta)(g) = \sigma(\theta(g))=\sigma(e_H)=e_K \ \Longrightarrow\ g\in\ker (\sigma\theta)$$
Then

$$\ker (\theta) \leq \ker(\sigma\theta)$$
  In particular, $|\ker (\theta)|$ divides $|\ker(\sigma\theta)|$.
  
  The other way around may not be true. There may be elements $g\in G$ such that $\theta(g) = h\neq e_H$ whereas $(\sigma\theta)(g)=\sigma(h)=e_K$ (if $h\in \ker\sigma$). For example, take $G=\mathbb{Z},\ H=\mathbb{Z}_4,\ K=\mathbb{Z}_2$ with $\theta$ and $\sigma$ the canonical projections. Then
  $$\ker(\theta) = 4\mathbb{Z}\qquad \ker(\sigma\theta) = 2\mathbb{Z}$$


*

*The First Isomorphism Theorem allows us to write:
$$G/\ker (\theta) \simeq \text{Im}\,(\theta) \qquad G/\ker (\sigma\theta) \simeq \text{Im}\,(\sigma\theta) $$
In particular, if $\theta$ and $\sigma$ are onto, then $\sigma\theta$ is also onto and:
$$G/\ker (\theta) \simeq  H \qquad G/\ker (\sigma\theta) \simeq K $$
And therefore
$$
|H| = \left\vert G/\ker (\theta)\right\vert = \frac{|G|}{|\ker (\theta)|} \qquad 
|K| = \left\vert G/\ker (\sigma\theta)\right\vert = \frac{|G|}{|\ker (\sigma\theta)|} $$
Which implies
$$|G| = |H||\ker (\theta)| = |K| |\ker (\sigma\theta)|\quad \Longrightarrow \quad
\frac{|\ker (\sigma\theta)|}{|\ker (\theta)|}=\frac{|H|}{|K|}$$
Therefore

$$\big[\ker (\sigma\theta) : \ker (\theta)\big]= \frac{|H|}{|K|}$$
  

A: We have
$$\mathrm{Ker}(\sigma\theta)=\{g|\sigma(\theta(g))=e\}$$
Since $\sigma(e)=e$ and $\theta(g)=e$ if $g\in\mathrm{Ker}(\theta)$, we have that $\sigma(\theta(g))=e$ if $g\in\mathrm{Ker}(\theta)$, so $g\in\mathrm{Ker}(\sigma\theta)$ if $g\in\mathrm{Ker}(\theta)$. Thus $\mathrm{Ker}(\theta)\subseteq\mathrm{Ker}(\sigma\theta)$.
For the last part, since $\sigma\theta:G\to K$ is surjective we have $|\mathrm{Ker}(\sigma\theta)|=|G|/|K|$. Also, $|\mathrm{Ker}(\theta)|=|G|/|H|$, so $|\mathrm{Ker}(\sigma\theta)|/|\mathrm{Ker}(\theta)|=|H|/|K|$.
