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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?

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

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No, your reasoning is not correct. For example, if there were two subgroups of order $5$, they can only intersect at $e$, the identity element or they are both the same subgroup.. This is because each subgroup has prime order and any intersection of subgroups is a subgroup. So, if there are two distinct subgroups of order $5$, the original group would have to have at least $25$ elements.

In fact for a group of order $10$, it can only have one subgroup of order $2$ and one of order $5$. Let $G$ be the group of order $10$, and $H$ the subgroup of order $5$. Note that $[G:H]=2$. It's not hard to show that any subgroup of index $2$ is normal.

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