Minimum size group with an element that is its own inverse

I Believe I saw a proof to the effect that groups of a minimum order were guaranteed to have at least one non-identity element that is its own inverse. I believe that the order was 24, but I cannot find said paper any more. Is this in fact true and can you provide a proof of this fact or refer me to a proof of this fact?

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Do you mean a non-identity element that is its own inverse? Then you are asking when is a group guaranteed to have idempotent. Many groups do not have these, even of very high order. – Lepidopterist Mar 16 '13 at 20:42
In fact, any odd order group cannot have such an element. – EuYu Mar 16 '13 at 20:43
An idempotent is not its own inverse. But an element of order 2 will be a nonidentity element which is its own inverse. I think any group of even order has one of these: $p|o(G)$ implies $G$ has an element of order $p$. – coffeemath Mar 16 '13 at 20:46
what are you trying to use this for? Maybe if we know the reason we can help you reformulate this question in a more helpful way? – Alexander Gruber Mar 16 '13 at 21:16
The overachring purpose is a proof that every group that has only trivial subgroups is finite and prime. Showing that for finite groups is trivial. I was trying to show that there is at least one subgroup in every infinite group. Our class has not dealt with infinite groups in much length so I was trying to se if their would be at least on non identity element that is its own inverse. But I see that infinite groups have an infinite number of subgroups. This however is not imeadiatly apparent to me. But it is sufficient to complete the proof. – SaintMacintosh Mar 16 '13 at 21:38

You are perhaps thinking of the note What is special about the divisors of $24$? where it is shown that if $n$ is a natural number such that every invertible element in ${\bf Z}/n{\bf Z}$ is its own inverse (or, equivalently, the group of units $U(n):=({\bf Z}/n{\bf Z})^\times$ has exponent $2$), then $n\mid24$. The converse holds true as well: if $n\mid24$, then $U(n)$ has exponent $2$. This simple, elementary fact may underlie a lot of deeper stuff "related to number theory, like lattices, moonshine, modular forms, string theory etc."
Another fun fact is that finite subgroups of ${\rm GL}_2({\bf Q})$ have order dividing $24$. This is a special case of theory worked out by Minkowski. See Part II of Lorenz's talk. – anon Mar 16 '13 at 21:44
Every group has an element that is its own inverse, namely the identity. If you mean a non-identity self-inverse element, then this is the same thing as an element of order $2$. A finite group $G$ has an element of order $2$ if and only if $|G|$ is even. Thus there is no result of the sort you want.