Existence of a normal subgroup in G

Question

Today on my algebra test I had such an exercise:

Let $|G|=66$. Show that there is a normal subgroup in $G$ of order $3$.

I am not even sure that's true. I wanted to show that $n_{3}$=1. But from what we know $n_{3}=1$ or $n_{3}=22$. I also get that $n_{11}=1$, so there is a normal subgroup of order 11. And a few possibilities for $n_{2}$.

Does anybody know if the statement above is correct and how to show it?

Answer 1

Hints:

1. By Sylow, there exists one unique Sylow $\;11$- subgroup $\;N\;$ , and thus $\;N\lhd G\;$

2. Let $\;P\;$ be any Sylow $\;3$-subgroup, so that $\;NP\lhd G\;$

3. There exists one unique group,up to isomorphism, of order $\;33\;$, which then thas a unique (and normal, of course) subgroup of order $\;3\;$ .

Deduce the claim now.

Further hint: If $\;P\lhd K\lhd H\;$ and $\;K\;$ cyclic, then $\;P\lhd H\;$

Answer 2

It is true.

Let $H$ be subgroup of order $3$ and $K$ be a subgroup of order $11$. Since $K$ is normal as you said $HK$ is a subgroup of $G$ with order $33$.

Notice that $H$ is normal in $HK$,(just apply sylow theorem for a group of order $33$) which means $H$ is uniqe in $HK$ as a sylow $3$ subgroup.

As $|G:HK|=2$ then $HK$ is normal subgroup of $G$ thus $H^g\leq (HK)^g=HK$ for all $g$ and since $H$ is uniqe we have $H=H^g$ for all $g\in G$.

Note: $X^g=gXg^{-1}$ as a notation.