Suppose that G is a finite group of even order. Show that G must contain an element $a\ne e$ for which $a^2 = e$

Question

Suppose that $G$ is a finite group of even order. Show that $G$ must contain a non- identity element a for which $a^2 = e$.

2

Think about inverses. – Angina Seng Oct 01 '17 at 14:47

score 1 · Answer 1 · answered Oct 01 '17 at 14:49

1

Hint. Consider the collection of sets $\{g,g^{-1}\}$ with $g\in G$. Note that $a^2=e$ is equivalent to $a=a^{-1}$.

answered Oct 01 '17 at 14:49

Robert Z

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I am thinking about inverses and whether the number of elements that are different from their own inverses are even or odd but I'm having trouble putting it down on paper. Just need a more solid starting point I feel frustrated with it. – Sammy.d Oct 01 '17 at 14:52
@Sammy.d That collection of sets gives you a partition of $G$. Such sets can have cardinality $1$ or $2$. – Robert Z Oct 01 '17 at 14:54

score 1 · Answer 2 · answered Oct 01 '17 at 16:08

Since the group is finite having even order, the number of non identity elements is odd (excluding identity element).

Now every one of these non identity elements have a unique inverse . As the number of non identity elements are odd, one non identity element will be left out at the end after pairing up the elements and their inverses.

Since this last element cannot have it's inverse same as that of any other element (it is a group), the inverse of that element will be itself ( that is a = a inverse).

Suppose that G is a finite group of even order. Show that G must contain an element $a\ne e$ for which $a^2 = e$

2 Answers2