Group of order $54$ has normal sugroup of order $27.$

Question

Let $G$ be a group of order $54$. Prove that there exists a normal subgroup of order $27.$ Is this normal subgroup unique?

Thoughts. Since $27$ divides $54$, by Lagrange's theorem we can not exclude the existence of such a subgroup.

Thank you in advance!

Answer 1

First Lagrange's theorem says that if $G$ has a subgroup $H$ then order of $H$ divides order of $G$. But converse may not hold.

In your case $|G|=54=3^3\times2 $. Now by applying Sylow theorem, we see that there is a Sylow-$3$ subgroup of $G$ having order $27$, say $H$. Also, $n_p$, the number of such a subgroup satisfies the conditions, $n_p|o(G)$ and $n_p\equiv1\pmod3.$ In this case only possible value for $n_p$ is $1$. Hence $H$ is a unique Sylow-$3$ subgroup. So $G$ has a normal subgroup of order $27$. (Unique Sylow-$p$ subgroups are normal.)

(This answer assumes the knowledge of Sylow theorems.)

Also note that one can use the fact that a subgroup of index $2$ is normal. Which makes things more simpler.

Answer 2

suppose that G is a group with order 2n where n is odd we can prove that G has a normal subgroup with order n .we know that G is isomorphic with H that H is a subgroup of symmetric group with order 2n now we have index of intersection of H and A2n in H less than index of A2n in S2n that equals to 2 since G has a member with order 2 (for example g) then H has a odd permutation then index of intersection of H and An in H is 2 now according to third teorm's isomorphic G is a subgroup with index 2 that is normal in G