If G is a finite group with an even number of elements, then binary product of two distinct elements is identity. Let $G$ be finite group, which has an even number of elements. Show that at least for two (distinct) elements $g,h$ of group $G$ one has $g*g = e$ and $h*h = e$.
I just started learning algebra and I have no ideas how I should solve this. I'm grateful for every explanation.
Reference: Fraleigh p. 48 Question 4.29 in A First Course in Abstract Algebra 
 A: Elementary way:  
For $g$ simply take the identity $e$. To find another, assume that each element $h$ has an inverse $h^{-1}$ that is not $h$ ($h \neq h^{-1}$). Summing the elements $\{h, h^{-1} \}$ and $e$ up, you get an odd number of elements of the group. Contradiction. So there is another element $h$ such that $h = h^{-1}$, and you are done.
Alternative: just refer to Cauchy's Theorem.
A: Hint: Separate the elements in the group into two sets: $A=\{x\in G\mid x^2=e\}$ and $B=\{x\in G\mid x^2\ne e\}$. Now, show that if $y\in B$ then $y^{-1}\in B$ and $y^{-1}\ne y$. What does that tell you about the parity of the number of elements in $B$? And since $|G|$ is even, what can you conclude about the parity of the number of elements in $A$? You now have your answer. 
A: Consider the relation on $G$ given by $g\equiv h\iff g\in\{\ h\ , h^{-1}\ \}$. It is easy to see that this is symmetric, reflexive, and transitive, and so an equivalence relation with equivalence classes $\{\ h\ ,\ h^{-1}\ \}$. The equivalence class of the identity $e$ of $G$ is $\{\ e\ \}$ containing only one element, and all equivalence classes have at most two elements. Since the order of $G$ is even, at least one equivalence class besides $\{\ e\ \}$ must have only one element, and that element is its own inverse. 
A: As an alternative way of looking at it:
Define an action of $G$ on itself by sending an element to its inverse. From class-equation we conclude that $|G|=|K|+\Sigma_i|O_i|$, where $K$ is the set of fixed points, and $O_i$ are orbits. Now each $l_i:=|O_i|$ cannot be one, and is in fact $=2$. Therefore, $2$ divides $|K|$, while $e\in K$, and thus the result follows.
Of course this is quite untidy, and the action here is just another way of saying "pairing". I think that this could make the answer look better however.
A: Proof 1 based on  http://answers.yahoo.com/question/index?qid=20080225171055AAJirM5. 
In any finite group, if an element (suppose $g$) isn't the same as its inverse, then the element can be paired with their inverse ($g$ can be paired with $g^{-1}$) to form a pair. Hence the total number of elements that $\neq$ their inverses is even. 
Hence $|group|$ = number of elements that don't equal their inverses $+$ $\color{purple}{\text{number of elements that equal their inverse}}$
$\implies$ even = even (from the previous paragraph) + $\color{purple}{\text{Number of elements that equal their inverse}}$ $\implies$ $\color{purple}{\text{Number of elements that equal their inverse}} =$ even.
Can this even number of elements that equal their inverse be $0$? NO, because the identity is such an element. So there must be at least two such elements, the identity and one more.  QED.
Proof 2 is based on http://answers.yahoo.com/question/index?qid=20071120213339AAinCRX and http://answers.yahoo.com/question/index?qid=20071125194802AA5xsJa. 
First look at $e$. Because $e*e = e$, $e$ is an element that is always equal to its inverse. But we want to find an element that does NOT equal its inverse. So forget about $e$.
Hence look at $G - {e}$. Question presupposes $|G|$ is even. This $\iff G - {e}$ has an odd number of elements. Take any element $a \in G - {e}$. Pair $a$ with its inverse.
Possibility 1. If $a^{-1}$ happens to be ${a},$ then $a^{-1}=a$. We're done.
Possibility 2. If not, then multiply $a$ by its inverse, which is NOT $a$ in this possibility, and delete the pair.
Now we have $G - \{e\} - \{a, a^{-1}\}$ which is still an odd number of elements left.
Go through or repeat possibilities 1 and 2 with this set. 
$G$ is presupposed as a finite group. Hence after a finite number of steps of considering the 2 above possibilities, the worst case is having only one element left. Remember, we forgot about $e$ so this isn't $e$. In this worst case, we can only pair $a$ with itself as the inverse of $a$. Put in other words, for $a \neq e$, we're left with $a = a^{-1}.$ But this is just $\iff a*a = e $. QED.
Also look at http://answers.yahoo.com/question/index?qid=20070725075050AAtHrZg, http://answers.yahoo.com/question/index?qid=20101206093620AAMVZU2, http://answers.yahoo.com/question/index?qid=20110901095234AAla6t0
