Show $H$ is normal in $G$

Question

$G$ has order $14$ and is not abelian. By Cauchy's theorem we know that it has an element $x$ of order $2$ and an element $y$ of order $7$.

I need to show that $H=\langle y\rangle $ is a normal subgroup in $G$. This is what I have done but I am not sure whether it is the correct approach:

$|H| = 7$, then number of Sylow $7$-subgroups $n_7 \equiv 1$ ($\text{mod } 7 = 1,8,... )$

but $n_7|2 \Rightarrow n_7=1 \Rightarrow$ number of Sylow $7$-subgroups $n_7= 1$

Thus there is an unique 7-subgroup and thus is normal.

Your proof is correct, I edited some sloppy use of notation. And, you really do not need Sylow theory to crack this problem. See @abdcef (what a name!). — Nicky Hekster, Dec 28 '17 at 10:37

score 2 · Accepted Answer · edited Dec 28 '17 at 10:07

2

It is a general fact that if $H$ is a subgroup of $G$, for finite groups with $2|H|=|G|$. Then $H$ is normal in $G$. See here for a simple proof.

edited Dec 28 '17 at 10:07

answered Dec 28 '17 at 10:04

abcdef

He asked: is my approach right? – 1ENİGMA1 Dec 28 '17 at 10:15
Sorry you're right, I overlooked the question. – abcdef Dec 28 '17 at 10:25
@1ENİGMA1 : This is discussed recently here. In particular take a look at Kevin's answer. – Dec 28 '17 at 10:46

1 Answers1