Prove if $G$ is a finite group and $H\unlhd G$ that there is a composition series of $G$ one of whose terms is $H$ 
Prove if $G$ is a finite group and $H\trianglelefteq G$ that there is a composition series of $G$ one of whose terms is $H$

So I previously proved every finite group has a composition series. Which is maybe useful here.
So I use two cases, either $G/H$ is simple or it is not. If $G/H$ is simple, then since $H$ has a composition series as it is a finite group, then $1\trianglelefteq H_1\trianglelefteq ...\trianglelefteq H\trianglelefteq G$ is a composition series of $G$ containing $H$.
If $G/H$ is not simple, then there exists some $K/H$ in $G/H$ such that $K/H\trianglelefteq G/H$. By correspondence theorem there exists a $K\trianglelefteq G$ containing $H$. I believe since $G$ is finite, I can also say that I can choose $K/H$ to have minimal index in $G/H$, since there can at most be finitely many $K_i/H\trianglelefteq G/H$ so there exists a minimal index. This ensures that $G/K$ is simple as if not then there exists an $N\trianglelefteq G$ containing $K$ which has $\vert G:N\vert\vert N:K\vert=\vert G:K\vert=\vert G/H:K/H\vert$, so then $\vert G/H:N/H \vert=\vert G:N\vert<\vert G:K\vert$ a contradiction.
Then since $H$ has a composition series, we get that $1\trianglelefteq H_1\trianglelefteq...\trianglelefteq H\trianglelefteq K\trianglelefteq G$ is a compositions series of $G$.
I think what I'm missing here however is that $K/H$ needs to be simple and I haven't shown that, I've only got that $G/K$ is simple. I don't for instance see anything wrong with $K/H$ not being simple, since I would get some $N$ with the property that $H\trianglelefteq N\trianglelefteq K$ where $K/N$ is simple but $N/H$ may not be and this would repeat again. And I believe since the group is finite this must terminate at some point.
 A: Okay, you are getting tied up in knots, and comment thread is getting too long.
Let $G$ be a group. By “a composition series” I understand a subnormal series
$$ 1=N_0\triangleleft N_1\triangleleft N_2\triangleleft\cdots \triangleleft N_{m-1}\triangleleft N_m=G$$
such that for all $i=0,\ldots,m-1$, $N_{i+1}/N_{i}$ is simple.
You say you already know that every finite group has a composition series.
Let $G$ be a finite group, and $H\triangleleft G$. We want a composition series for $G$ with $N_i=H$ for some $i$.
Since $H$ is finite, $H$ has a composition series,
$$1=N_0\triangleleft N_1\triangleleft\cdots \triangleleft N_r=H.$$
Now, $G/H$ is also finite, so it has a composition series, which for ease I will number starting at $r$:
$$ 1 = M_r\triangleleft M_{r+1}\triangleleft \cdots \triangleleft M_{r+s}=\frac{G}{H}.$$
By the correspondence theorem, each $M_i$ corresponds to a subgroup $N_i\leq G$ (I will show shortly that the “new” $N_r$ is the same as the old one) that contains $H$, and moreover, $N_i\triangleleft N_{i+1}$ because the correspondence preserves normality. Note that $N_r/H = M_r=\{1_{G/H}\}$, so both subgroups called $N_r$ are actually equal and I’m not abusing notation. Now we have a subnormal series
$$1 = N_0\triangleleft\cdots\triangleleft N_r =H \triangleleft N_{r+1}\triangleleft\cdots \triangleleft N_{r+s}=G.$$
It only remains to show this is a composition series.
For $i=0,\ldots,r-1$, the quotient $N_{i+1}/N_i$ is simple, since the $N_i$ are a composition series for $H$. For $i=r,\ldots,r+s-1$, we have
$$\frac{N_{i+1}}{N_i} \cong \frac{N_{i+1}/H}{N_i/H} \cong \frac{M_{i+1}}{M_i}.$$
The first isomorphism follows from the Isomorphism Theorems (usually second or third, depending on how you number them, which tells you that a quotient of a quotient is a quotient of the original group).
Now, we know $M_{i+1}/M_i$ is simple, because the $M_i$ form a composition series for $G/H$. Therefore, $N_{i+1}/N_i$ is simple for $i=r,\ldots,r+s-1$ as well.
This shows that in our subnormal series, all quotients are simple. Thus, it is a composition series. And $N_r=H$, so it is a composition series one of whose terms is $H$, as desired. $\Box$
