Every element of a finite group to the power of the order of the Group is 1 - is this true - what is the proof for this?

Question

Going through Christof Paar's book on cryptography. In his chapter on DHKE, he has the following

The book doesn't seem to have the proof for $a^{|G|} = 1 $ Also how is it a generalization of Fermat's Little Theorem for Cyclic Groups?

Answer 1

You've worded the question in a way that suggests the book does have the proof of part 2. But, just in case, the proof is that any $a\in G$ generates a (cyclic) subgroup of $G$ of size $ord(a)$ and, moreover, Lagrange's Theorem implies that the size of any subgroup divides the size of the whole group. The proof Lagrange's Theorem is that if $H$ is a subgroup of $G$, then $G$ is partitioned into the left cosets of $H$, each of which has the same size as $H$, and so $|H|$ divides $|G|$.

Now part 1 follows immediately. If $m=ord(a)$ then $|G|=mn$ for some $n$ by 2. So $a^{|G|}=a^{mn}=(a^m)^n=1^n=1$.

Fermat's little theorem says that for any prime $p$ and any integer $a$ not divisible by $p$ we have $a^{p-1}\equiv 1$ (mod $p$). Since it is enough to consider the remainder of $a$ mod $p$, we may view $a$ as an element of the group $G=\{1,\ldots,p-1\}$ under multiplication mod $p$. Now Fermat says that $a^{p-1}=1$ for any $a\in G$. This is a special case of part 1 of your theorem.

Answer 2

Each translation $gH:=\{gh:h\in H\}$ of a $G$ subgroup $H$ is called a $\it{left}\;H\;coset$. The left $H$ cosets all have the same size $|H|$ and $$g'\in gH\implies\exists\;h\in H\;\;g'H=ghH=g(hH)=gH$$ Therefore $gH$ is the unique left coset containing $g$ and thus the family $(G:H):=\{gH:g\in G\}$ is a partition of $G$ into parts of equal size. $$\therefore\;|H|\;\Big\vert\;|G|=\sum_{X\in(G:H)}|X|\;\;\;\therefore\;\text{ord}(a)=|\langle a\rangle|\;\Big\vert\;|G|\;\;\;\therefore\;a^{|G|}\in\langle a^{\text{ord}(a)}\rangle=\langle\mathbf 1\rangle\;\;\;\therefore\;a^{|G|}=\mathbf 1$$