Prove that a group of order 25 has a subgroup of order 5 I found this question 
Subgroup(s) of a group of order 25 
I want to know if proving such a statement is possible by contradiction. 
Question: Let G be a group of order 25. Prove that G has at least one subgroup of order 5.
 A: First, note that $G$ has a subgroup of order $5$ if and only if $G$ has an element of order $5$. Let's quickly prove this fact.
If $G$ has a subgroup $H$ of order $5$, and $g$ is any nonidentity element of $H$, then $\langle g \rangle$ is a subgroup of $H$ with order greater than $1$. By Lagrange's theorem, $\langle g \rangle$ must have order $5$, so $g$ is an element with order $5$.
Conversely, if $G$ has an element $g$ of order $5$, then $\langle g\rangle$ is a subgroup of order $5$.
Now let's prove your desired result:

If $G$ has order $25$, then it has at least one subgroup of order $5$.

Proof by contradiction: By the remarks above, this is equivalent to proving that $G$ has an element of order $5$. Suppose for the sake of contradiction that $G$ has no element of order $5$. By Lagrange's theorem, every element must therefore have either order $1$ or order $25$. Only the identity has order $1$, so all other elements have order $25$. In particular, $G$ is cyclic, generated by some $g \in G$ with order $25$. But then $g^5$ has order $5$.
The same proof works for any prime $p$. Simply replace $5$ with $p$ and $25$ with $p^2$.

Note that we don't really need to use a proof by contradiction here. In the proof above, we assumed that something was false ($G$ has no subgroup of order $5$), then proved that it must be true ($G$ has a subgroup of order $5$ after all). We could just as well prove it directly, as follows:

Direct proof: If $|G| = 25$, then every element of $G$ has order $1$, $5$, or $25$ by Lagrange's theorem. Let $g$ be any nonidentity element. If $g$ has order $5$ then we're done. Otherwise, $g$ has order $25$, so $g^5$ has order $5$. In any case, $G$ has an element of order $5$.
