2

This is not for homework, and I've shown the problem without using this hint before, but I am just trying to understand how the following hint is helpful. The problem asks:

If $G$ is a finite group with even order, show that there is an element $a \neq e$ of $G$ such that $a = a^{-1}$. (Hint: Use the fact that $(a^{-1})^{-1} = a$ for every $a \in G$.)

I started by writing $G = \{e, a_1, \dotsc, a_m \}$. There must be some integer $i$ such that $a_i = a_1^{-1}$. From here I'm thinking there is some trick in manipulating the elements of $G$? maybe inverting every element? or multiplying every element by $a_1$?

tylerc0816
  • 4,123
  • Call two elements $x,y$ equivalent if $x = y$ or $x^{-1} = y$. Show that that is an equivalence relation. How many elements does each equivalence class have? – Daniel Fischer Sep 24 '13 at 23:51

2 Answers2

1

The map $x\mapsto x^{-1}$ is a bijection. It has one fixed point, $x=e$. The elements that are not fixed points come in pairs, because as in the hint, $(a^{-1})^{-1} = a$. Since $G$ has an even number of elements, there must be another fixed point, that is, $a \neq e$ such that $a = a^{-1}$.

In fact, there is an odd number of elements $a \neq e$ such that $a = a^{-1}$.

lhf
  • 216,483
1

An additional way to visualize this problem:

If you create the Cayley table (the group multiplication table) for $(G, *)$ then you will need to have exactly one $e$ appearing in each and every row. Since $|G|$ is even, the total number of $e$'s is the same as the total number of rows, hence the total number of $e$'s is even.

Note that whenever $xy = e$, we have $yx = e$ as well. (Is this familiar?) Therefore, each $x$ that has a different element $y$ as its inverse contributes two $e$'s to the table: one for $x*y$ and one for $y*x$.

Without any other element being its own inverse, this would contribute an even number of $e$'s to the table; along with the $e$ we get from $e*e$, this shows the total number of $e$'s is odd.

This gives a contradiction, for the number of $e$'s in the table cannot be both even and odd.

Benjamin Dickman
  • 14,451
  • +1 Very intuitive! I don't see how this uses the hint, though. – tylerc0816 Sep 25 '13 at 00:08
  • @tylerc0816 Yes, using the "hint" would feel a bit contrived here. I just thought I'd contribute another visualization for fun. If you find this response to be of interest, then you might enjoy this question: http://math.stackexchange.com/questions/221813/adapting-a-proof-on-elements-of-order-2-from-finite-groups-to-infinite-groups – Benjamin Dickman Sep 25 '13 at 00:13