If G is a finite group, and $\exists !$ $H\leq G$ such that $|H| =n$, then $H \triangleleft G$.

126 Views Asked by At

So far this is what I'm thinking:

Suppose the $G$ has exactly one subgroup of size $n$.

Let $N$ be a subgroup of $G$ of size $n$.

By Lagrange's Theorem we know $|N| \mid |G|$

But pretty much after that I'm a little confused about where to go and I'm not even sure I'm starting off correctly. I know to be a normal subgroup $N$ must be a subgroup of $G$ and if $a\in N$ and $b\in G$ then $b^{-1}ab\in N$. I do remember something that if $|N| = n,\;$ let $a\in G$ such that $N=\{a^{n}\mid a\in \mathbb{N}\}$ so $\,N\,$ is the subgroup generated by $\langle a \rangle,\,$ which means $\,O(a) \mid |G|\,$ and is possibly isomorphic to $\langle \mathbb{Z}_{o(a)},+,-,0\rangle$.

I'm really not sure and I'm pretty sure all I just said is wrong.

1

There are 1 best solutions below

7
On

A subgroup $H$ of $G$ is normal if $g^{-1}Hg=H$. Show that $g^{-1}Hg$ is always a subgroup of $G$ and that it has the same size as $H$ does. Conclude the result.