A simple question about normal subgroups of finite groups.

Question

Let $G$ be a finite group and let $H_1 \trianglelefteq G$. If $H_2 \leq G$, with $|H_1| = |H_2|$, can we state that $H_2 \trianglelefteq G$? If it is false in general, is the same statement true supposing furthermore that $H_1$ and $H_2$ are isomorphic?

Answer 1

No. Pick your favourite example of $A\le B$ with $A\not\lhd B$. Then consider $G=A\times B$ and let $H_1$ be the $A$ that is the first factor and $H_2$ the $A$ contained in the second.

Answer 2

For the first argument there is a well-known which results: If $$|G/H_1|=2$$ where $|G|<\infty$ so, $|G/H_2|=2$ so $H_2$ would be a normal subgroup. So this, as you stated, may not be true in general. Moreover, we could have the first statement true when $p$ be the smallest prime number which divides $|G|$ and $|G/H|=p$.

Answer 3

No. Take the group of all $2\times2$ matrix over $\mathbb{F}_3$ (or any other finite field whose characteristic is not $2$). Now, consider $H_1=\{\pm\operatorname{Id}\}$ and$$H_2=\left\{\operatorname{Id},\begin{bmatrix}1&0\\0&-1\end{bmatrix}\right\}.$$

Answer 4

It is true also if $H_1$ is a Sylow subgroup of $G$ (this is the Second Sylow theorem).

Answer 5

Several subgroups of the symmetric group $S_4$ are isomorphic to the Klein Four-Group, but only one of them is a normal subgroup.