202004240713 Homework 1 (Q3)

Suppose $G$ be a finite group of even order. Prove that there exists an element $a\in G$ such that $a\neq e$ and $a^2=e$ .

Attempts.

Try negating the statement by the following:

$\forall\, a\in G$ , either $a=e$ or $a^2\neq e$ ,

then looking for contradiction to the assumption that $G$ should be a finite group of even order.

CASE 1: Would it be possible that $a=e$ ?

Given so, the inverse and the identity of $G$ would be $e$ . And the set $G$ would have contained $e$ only, i.e., the singleton set $G=\{ e\}$ . This contradicts with the assumption that its order be even.

CASE 2: If for all elements in the group $G$ , their order cannot ever be $2$ .

If $a^2\neq e$ , then $a\neq a^{-1}$ for all $a\in G$ . It would then become an ill-posited negation, because the identity element $e$ is one of all $a\in G$ , and both statements $e^2\neq e$ and $e\neq e^{-1}$ are abhorrent to the basic axioms of a group. So perhaps to my discretion $e$ should be precluded from $a$ ‘s, and understood as one ( $e=e^{-1}$ ) member apart from the other members.

Proceeding to observe that if $a\neq a^{-1}$ for all $a\in G$ , I may then construct a pair of $a_i$ and $a_i^{-1}$ for $i\in \{ 1,2,\dots ,n:n\in\mathbb{Z}^+\}$ , provided from the definition of a group that the inverse of each element must exist and, as it follows, that it must be unique. That is, if $a_5=a_{27}^{-1}$ , I may relabel $a_{27}$ as $a_5^{-1}$ . By such a construction I can guarantee that all elements, except the identity $e$ , are now in pairs.

One therefore, by counting in total how many elements there are in group $G$ , will get an odd number $(2n+1)$ , the $2n$ counted from the pairs, and the single $1$ the identity $e$ itself alone counts.

This contradicts with the assumption that $|G|$ be of even parity.

From the negation of statement arises contradiction, the original statement is therefore proven by contradiction.

Remark.

I could not prove otherwise than loosely to so naive myself. Not until I had survived mathematical rigor.

M	T	W	T	F	S	S
		1	2	3	4	5
6	7	8	9	10	11	12
13	14	15	16	17	18	19
20	21	22	23	24	25	26
27	28	29	30

CASE 1: Would it be possible that ?

CASE 2: If for all elements in the group , their order cannot ever be .

Leave a Reply Cancel reply

CASE 1: Would it be possible that $a=e$ ?

CASE 2: If for all elements in the group $G$ , their order cannot ever be $2$ .