Htykuut

Let $G$ be a finite group, and the order of the group $G$, $|G|=r$.

For any $x in G$, prove that $x^{-r}=e$ (where $e$ is a neutral in $G$).

Basically, I haven't tried anything except trying to manipulate the expression $aa^{-1}=e$ into what I'm trying to prove because I really have no idea what to do.

I would appreciate any hint, advice or etc. because I'm new in group theory and the concepts are still hard to understand.

Also, the next part of this task is this:

$G$ is a group such that $|G|=195$. Is there an element $g neq e$ such that $g^{77}=e$?

Thanks in advance!

Suppose $o(x)=d$. This is the same as saying that $o(langle x rangle)=d$. By Lagrange's Theorem (which I assume you know), $d$ divides $r$, so $r/d$ is an integer, call it $m$. Then $x^r = (x^d)^m=e^m=e$.

For the second question: if there is such a $g$, then since $g^{77}=e$, the order of $g$ must be a divisor of $77$ (do you see why?). But, again by Lagrange, the order of $g$ must also divide $195 = |G|$, thus it must also divide $gcd(195,77)=1$. The only element of order $1$ in any group is $e$, which is a contradiction to the assumption that $g neq e$.

If $g^{77}=1$, then $(g^{77})^2=g^{154}=1$ and $(g^{77})^3=g^{231}=g^{231(mod 195)}=g^{36}=1$.

Notice $195=3*5*13$ and $77=7*11$ so $lcm(77,195)=77*195$ which means $77*a neq 0(mod195)$ for $a<195$. So by applying the above argument, $(g^{77})^{38}=g=1$. So contradiction.