# Thread: Finite groups of even cardinality

1. ## Finite groups of even cardinality

Let (G,*) be a finite group with an even number of elements. Show that there must exist at least one element $\displaystyle a \neq e \in G$ such that $\displaystyle a^2=e$.

I have no clue how to start and what ticks me off about that is that I think I've seen it before. (And I don't recall it was that difficult, either.) However my class notes (from years ago) doesn't contain the proof.

-Dan

2. Consider pairing up elements with their inverses.

3. Originally Posted by rgep
Consider pairing up elements with their inverses.
Got it. Let me formalize it then.

Let (G,*) be a finite group with an even number of elements. Consider the subset of G: $\displaystyle A \equiv \{b|b^2=e, b \in G\}$ where e is the identity in G. It may be that A is empty. If not, the inverse of any element $\displaystyle b \in A$ is also in A and is distinct from b. So b and its inverse form a pair of elements in A and this is true of any element in A. Thus we know that A has no elements or has an even number of them.

Now form the set G - A. (For others, I believe the notation is G\A?) We know that G - A is not empty because $\displaystyle e^2=e$. We also know that the set G - A has an even number of elements since both G and A do (calling 0 an even number). Because of this, the set G - A - {e} cannot be empty, since this would imply that G has an odd number of elements. Since any element $\displaystyle a \in G-A-{e}$ is not in A we know that $\displaystyle a^2=e$.

Thus there is at least one element a in G that is not the identity that has the property $\displaystyle a^2=e$. (End of proof.)

Wow. That took a lot more writing than I thought it would.

Thanks for the tip!

-Dan

4. Originally Posted by topsquark
Got it. Let me formalize it then.

Let (G,*) be a finite group with an even number of elements. Consider the subset of G: $\displaystyle A \equiv \{b|b^2=e, b \in G\}$
Don't you need to add $\displaystyle b\ne e$ to the definition of $\displaystyle A$?

where e is the identity in G. It may be that A is empty. If not, the inverse of any element $\displaystyle b \in A$ is also in A and is distinct from b.
Take the group $\displaystyle \mathcal{G}=(\{e,b\},*)$, where $\displaystyle b=b^{-1}$. Then $\displaystyle b \in A$ but $\displaystyle b^{-1}$ is not distinct form $\displaystyle b$.

So b and its inverse form a pair of elements in A and this is true of any element in A. Thus we know that A has no elements or has an even number of them.

Now form the set G - A. (For others, I believe the notation is G\A?) We know that G - A is not empty because $\displaystyle e^2=e$. We also know that the set G - A has an even number of elements since both G and A do (calling 0 an even number). Because of this, the set G - A - {e} cannot be empty, since this would imply that G has an odd number of elements. Since any element $\displaystyle a \in G-A-{e}$ is not in A we know that $\displaystyle a^2=e$.

Thus there is at least one element a in G that is not the identity that has the property $\displaystyle a^2=e$. (End of proof.)

Wow. That took a lot more writing than I thought it would.

Thanks for the tip!

-Dan

5. Originally Posted by topsquark
Let (G,*) be a finite group with an even number of elements. Show that there must exist at least one element $\displaystyle a \neq e \in G$ such that $\displaystyle a^2=e$.

I have no clue how to start and what ticks me off about that is that I think I've seen it before. (And I don't recall it was that difficult, either.) However my class notes (from years ago) doesn't contain the proof.

-Dan
If there is no such element every element $\displaystyle a$ except $\displaystyle e$ has a distinct
partner $\displaystyle a^{-1}$, adding up to an even number of elements, but
$\displaystyle e$ is un-partnered so there are an odd number of elements in the