# 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.

On a side note, if you are trying to avoid cosets because they seem frighteningly confusing, I have good and bad news. The bad news is that they are simply unavoidable if one wants to achieve any degree of competency in group theory. But the good news is that often students find cosets quite confusing at first, but eventually after enough examples and time spent thinking about them, many students become proficient in using them.

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