# Finding all quotient groups $F/N$ up to isomorphism where $N \leqslant F \leqslant \langle H, N \rangle$.

Suppose $H$ is a subgroup of a group $G$, where $|H|=p$, for $p$ a prime. I need to describe, up to isomorphism, all possible quotient groups $F/N$, where $N \trianglelefteq G$ and $N \leqslant F \leqslant \langle H, N \rangle$.

Here is my attempt thus far - note that it includes a lot of subcases and sub-subcases, so if it turns out I don't need to make it so complicated, please let me know how to do so.

Since $\langle H, N \rangle$ is the smallest subgroup containing both $H$ and $N$, and $N \leqslant F$, we have three cases:

1. Case (i): Neither $H$ nor $\langle H,N\rangle$ coincides with $F$.

In this case, $F \leqslant H$, and by Lagrange's Theorem, $|F|$ must be a divisor of $|H|$. However $|H|=p$, so either $|F|=p$ or $|F|=1$.

• Subcase(i): $|F|=1$. Then, $F$ is the trivial subgroup $\{ e\}$, so $F/N \simeq N$, and since $N \leqslant F,$ we have that $N \leqslant \{e\},$ which implies that $N$ is also $\{e\}$, So, $F/N\simeq \{e\}$.
• Subcase (ii): $|F|=p$. Then, by Lagrange's Theorem, either $|N|=p$ or $|N|=1$.

• Subsubcase (i): $|N|=1$, then $N$ is the trivial subgroup $\{e\}$, so $F/N \simeq F \simeq \mathbb{Z}_{p}$, since $|F|=p$ and every group of prime order is isomorphic to $\mathbb{Z}_{p}$.

• Subsubcase (ii): $|N|=p$. Then, since $|F/N|=|F:N|$, and by Lagrange's Theorem, $|F|=|N||F:N|$, we have that $\displaystyle |F|=|N||F/N| \implies \frac{|F|}{|N|} = |F/N| \implies 1=|F/N|$. So $F/N \simeq \{e\}\\$

1. Case (ii): $H$ coincides with $F$

In this case, $|H|=|F| = p$, so by Lagrange's Theorem, we have the following two subcases:

• Subcase(i): $|N|=1$. Then, $N$ is the trivial subgroup $\{ e\}$, so $F/N \simeq F \simeq \mathbb{Z}_{p}$.
• Subcase (ii): $|N|=p$. Then, by Lagrange's Theorem, $\displaystyle \frac{|F|}{|N|}=1$, so $F/N \simeq \{e\}$.

1. Case (iii): $F$ coincides with $\langle H, N \rangle$

Then, $N \leqslant H \leqslant F \simeq \langle H,N \rangle$, and we have two subcases:

• Subcase(i): $|N|=1$. Then, $N$ is the trivial subgroup $\{ e\}$, so $F/N \simeq \langle H, N \rangle /N \simeq HN/N \simeq HN/\{e\}$ (since $\langle H,N\rangle$ is defined to be the subgroup $HN$ of $G$ generated by the union of $H$ and $N$). But, by the Second Isomorphism Theorem, $H/(H \cap N) \simeq HN/N$, and if $N$ is the trivial subgroup $\{e\}$, then $H\cap N \simeq \{e\}$, so $HN/\{e\} \simeq H/(H \cap \{e\}) \simeq H/\{e\} \simeq H$, and since $|H|=p$, in this case, $F/N \simeq \mathbb{Z}_{p}$.

• Subcase (ii): $|N|=p$. Then, $N$ must boincide with $H$. So, by Lagrange's Theorem, $\displaystyle |F|=|N||F/N| \implies \frac{|F|}{|N|}=|F/N|$. So, we have two sub-subcases:

• Subsubcase (i): $|F|=p$, then $N$ and $H$ must coincide with $F$ and with $\langle H, N \rangle$. So, by Lagrange's Theorem, $\displaystyle|F/N|=\frac{|F|}{|N|} = \frac{p}{p} = 1$. So $F/N$ is the trivial subgroup $\{e\}$.

• Subsubcase (ii): $|F|>p$. Then, this mmeans that $|F| = kp$, where $k \in \mathbb{N}$, $k > 1$. Then, by Lagrange's Theorem, $\displaystyle |F/N| = \frac{|F|}{|N|} = \frac{kp}{p} = k$.

Now, from this point, I don't know where to go. There's no guarantee that $F/N$ is cyclic, so I can't say it's $\simeq \mathbb{Z}_{k}$. So, there really isn't anything specific I can say about $F/N$ in this case, which I'm sure I'm supposed to be able to, which means I must have done something wrong.

I'm really quite stuck at this point, and desperately need help to finish this problem!

Thanks.

• Is $G$ a finite group? Oct 23 '16 at 0:16
• @Berci we don't know that.
– user100463
Oct 23 '16 at 0:17
• There are some uncertain points in your work: 1. Why $F\le H$ in this case? Similarly, at 3. How $N\le H$ could follow? Oct 23 '16 at 0:19
• @Berci $F \leqslant H$ because $\langle H, N \rangle$ is the smallest subgroup containing both $H$ and $N$, so if $N \leqslant F$ and $F$ is not $H$ or $\langle H, N \rangle$, then we must have $F \leq H$. For 3., that was given.
– user100463
Oct 23 '16 at 0:28

Hints:

1. $\langle H,N\rangle = HN$ since $N$ is normal subgroup
2. $N\le F\le HN$ implies ${\bf1}\le F/N \le HN/N$
3. $|HN/N|\le |H|$.
• I actually say 1. in my proof. I have never heard of 2. and 3. before. Also, how do those things fit into the framework of my proof?
– user100463
Oct 23 '16 at 0:26
• I'm not sure this really counts as an answer...because you don't really say how any of those three hints are relevant.
– user100463
Oct 23 '16 at 0:29
• They might not fit in the framework of your proof, but they are hints towards a much easier proof. The proof should only take about 2 or 3 lines, and this gives you most of it. If you have never seen 3, look up the second isomorphism theorem. For 2, see en.wikipedia.org/wiki/Correspondence_theorem_(group_theory) Oct 23 '16 at 8:10
• @verret the Second Isomorphism Theorem tells us that if $H \leqslant G$ and $N$ is a normal subgroup of $G$, then $H/(H \cap N) \simeq HN/N$. How does that tell us that $|HN/N|\leq |H|$? By Lagrange's Theorem? Because then, $\displaystyle |H/(H \cap N)| = \frac{|H|}{|H \cap N|}$?
– user100463
Oct 24 '16 at 3:23
• @JessyCat yes... Oct 24 '16 at 18:38