Prove that a group of order 5 doesnt have any self-inverse member other than the identity member. Assuming you only know the most basic ideas about groups properties-
If group G contains 5 members, how do you prove that no member other than the identity member is self inverse?
*I rather have a clue on how to begin solving it or how to approach this problem than have the whole proof :)

Assume that $a=a^{-1}$ and $a \neq e$. $G$ can be partioned in two disjoint sets: put $S = \{x \in G: x=x^{-1}\}$ and $T=\{x \in G: x \neq x^{-1}\}$, then $G = S \cup T$ and since $a, e \in S$, $|S| \geq 2$. 
Note that $|T|$ is even (elements of $T$ come in pairs), which leaves us with $|T|=0$ or $2$ (if its cardinality would be equal to $4$ then $5 \geq 2+4$, which is absurd).
If $T$ is empty then every element of $G$ equals its inverse. Next to $e$ and $a$, let $b$ a third element of $G$ different from $e$ and $a$. Then clearly, $ab \in G$ and $ab \notin \{e,a,b\}$. Note that $ab=(ab)^{-1}=b^{-1}a^{-1}=ba$. So $G=\{e,a,b,ab,c\}$ for some $c$ not equal to any of the other elements. Since $ac \in G$ we now reach a contradiction: go verify that $ac \neq e, ac \neq a, ac \neq b, ac \neq ab, ac \neq c$.
If $|T|=2$, then $S=\{1,a,b\}$ and $T=\{c, c^{-1}\}$, for certain $b,c \in G, a \neq b \neq e$. Since $G$ is closed, $ab \in G$ and it is trivial to see that $ab \notin \{e,a,b\}$. Hence (after appropriate renaming) we can assume $ab=c$ and thus $(ab)^{-1}=b^{-1}a^{-1}=ba=c^{-1}$. 
So after all, $G=\{e,a,b,ab, ba\}$. Now for the final contradiction, we must have that $bab \in G$, but go verify it is not equal to any of the elements of $G$ (for example, if $bab=e$, then $ba=b^{-1}=b$, whence $a=e$, etc..)
