A group of order $16$ with a normal subgroup of order $4$.

Question

This is an exercise assigned by my prof, for which I need some help. If $G$ is a group of order $16$ with a normal subgroup $N$ of order $4$, why for every $a, b \in G$ we have that $ab=ban$ for some $n \in N$ ?

The order of the quotient is $4$, so is abelian, then if $a, b \in G$ we have that $aNbN=bNaN$ and then?

Hint: you can say something pretty definite about the quotient. — Randall, Apr 05 '22 at 17:56
Hint: every group of order $4$ is abelian. But please show some effort of your own, otherwise the community will close this post. — Nicky Hekster, Apr 05 '22 at 17:59
so $abN=baN$ and for every $n_1, n_2 \in N$ we have $abn_1=ban_2$ thus $ab=ban_2n_1^{-1}$? — Rick88, Apr 05 '22 at 18:08

Shaun · Answer 1 · 2022-04-05T20:25:07.683

1

By Lagrange's Theorem, $|G/N|=|G|/|N|=16/4=4$. Every group of order four is abelian. Therefore,

$$\begin{align} (ab)N&=(aN)(bN)\\ &=(bN)(aN)\\ &=(ba)N \end{align}$$

for all $a,b\in G$. But this means $ab=(ab)e\in (ab)N=(ba)N$ (because $e\in N$ since $N\unlhd G$), which implies there exists an $n\in N$ with $ab=ban$.

edited Apr 05 '22 at 20:25

answered Apr 05 '22 at 19:15

Shaun

44,997

A group of order $16$ with a normal subgroup of order $4$.

1 Answers1