I have a question about groups of finite order.

Question

I want to show that if a group G has finite even order then it must have an odd number of elements that are their own inverse, and if G has odd order then it has no elements of order two. I know this is trivial with lagrange's theorem, but I don't want to use it. For the case when n is even, if we remove identity and all elements that are NOT their own inverse, we have an odd amount of elements remaining and each is its own inverse. I am not sure how to go about the case when n is odd. I tried the same approach, but after removing the elements that are not their own inverse and identity, we are left with an even amount, but I don't know how to show that number is 0.

Answer 1

Suppose $a \in G$ is an element of order 2. Let $H = \{e,a\}$ and for $g \in G$, $H_g=\{g,ga\}$. We show that if $H_g \neq H_h$, then $H_g \cap H_h =\emptyset$. Suppose for a contradiction that $x \in H_g \cap H_h$. Then with out any lost of generality, we can assumme that $x = g$ and $x =ha$. Which implies that $ g = ha$ and $ h = ga$, which is a contradiction. Now, for any $g \in G$, we have $g \in H_g$. Thus $G = \cup_{g \in G} H_g$, which implies that $|G| = 2m$ for some integer $m$ and $2 \mid |G|$, which is a contradiction. So $G$ has no element of order 2.

Answer 2

Suppose some element $a$ has order two. Then $a$ acts on $G$ by permuting all of the elements, and since $a^2b = b$ for all $b$ this permutations is a product of disjoint transpositions. Since $|G|$ is odd, $a$ fixes some element $b$, so $ab = b$ and $a = e$ which has order 1, a contradiction.

Answer 3

You probably want to know that the following more general fact holds true.

Theorem Let $G$ be a finite group and $p$ a prime dividing the order of $G$, then the number of elements of order $p$ of $G$ $\equiv -1$ mod $p$.

The proof can be easily derived from the proof of the theorem of Cauchy (existence of an element of order $p$) as given by James McKay.