[Math] All groups of order 10 have a proper normal subgroup

Question

Problem. Prove that if $G$ is a group of order 10, then $G$ contains a proper normal subgroup.

Our attempt. If $G$ is cyclic, then $G \cong \mathbb{Z}_{10}$, which is abelian, and thus any proper subgroup, e.g. $<2>$, is normal.

If $G$ not cyclic then, by Lagrange's theorem, any proper subgroup can have only order 2 or 5. This, together with the fact that G has 10 elements in total, generates three cases:

1. There are three proper subgroups. One of order 2 and two of order 5.

2. There are six proper subgroups. Five of order 2 and one of order 5.

3. There are nine proper subgroups, all with order 2.

(All these subgroups must be cyclic as they have a prime number of elements.)

Is this reasoning reasonable? Any ideas on where we can go from here?

Answer 1

By Cauchy's theorem, $G$ has a subgroup of order $5$. Since this subgroup has index $2$, it is automatically normal.