# Proof that the only proper subgroup of a group of prime order is the trivial subgroup without using cosets

Is it possible to prove the following without using cosets?

The only proper subgroup of a group of prime order is the trivial subgroup.

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–  joriki Oct 25 '12 at 17:41
Why do you want to do this? –  Chris Eagle Oct 25 '12 at 17:49

Consider a non-trivial subgroup $H$ of $G$. Let $x \in H$ be any element other than the identity. What is the order of $x$? Well, it must divide the order of $G$ which is prime and it is bigger than $1$ (because only the identity has order $1$). So the order of $x$ is equal to the order of $G$. Thus every element of $G$ is in $H$, since $x$ is in $H$ by hypothesis and $G$ consists of powers of $x$. So $H = G$.
No mention of cosets! But ... I cheated a bit. The key fact used is that the order of $x$ divides the order of $G$, and that is a corollary of Lagrange's Theorem (that the order of a subgroup divides the order of the group). And the way to prove Lagrange's Theorem is to use cosets.