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In preparation for a test, I have been looking at examples I covered in class and trying to understand them better. I recently did the following example:

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If $a^2$ does not equal e for all a, then a does not equal $a^{-1}$, hence this pairing exhausts all non-identity elements. Therefore the number of elements of the group is $2n + 1$, where n is the number of pairings and 1 is the identity element e. But we assumed that G was a group of even order, so this is a contradiction. So G contains an element of order 2.

Does this seem right?

Thanks for reading, greatly appreciate it!

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    Do you realize that both questions you copied and pasted above are **exactly one and the same question** ? – 2017-01-31
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    @DonAntonio how so? Is $a^2$ = $e$ exactly the same as $a = a^{-1}$ but just written differently? – 2017-01-31
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    $a^2 = e$ and multiply both sides by $a^{-1}$. – 2017-01-31
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    @jeremy Exactly so. – 2017-01-31
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    @DonAntonio Oh wow, I didn't know you could do a move such like that. So is it safe to say that $a = a^{-1}$ $<=>$ $a^2 = e$? Thanks Student! – 2017-01-31
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    @JeremyCho Yup, exactly so. An element $\;a\;$ in a group is an involution iff $\;a^2=e\;$ iff $\;a=a^{-1}\;$ , and $\;a\neq 1\;$ . – 2017-01-31
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    @DonAntonio Wow, thanks for the clarification! – 2017-01-31
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    Consider for example the cyclic group of order 6, that is $\{e, x, x^2, \ldots, x^5\}$. In this group we have that $x^6 = e$ and therefore $a^{-1}a^6 = a^{-1}e$ and adding exponents in the left hand side, we have that $a^5 = a^{-1}$... This to show that you can always multiply some equation by inverses (essentially the same what you do when you solve some equation in $\mahrbb{R}$ for example) – 2017-01-31
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    @DonAntonio not exactly the same question. The first question says if G contains a subgroup of order 2, the order of G is even. The next says that if G is even it has a subgroup of order 2. – 2017-01-31
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    @DougM I don't know what you're reading. In the **original question**, it was written what is written now and then another piece copied and pasted that said that if $\;G\;$ is a group of even order then *prove* there's an element $\;a\;$ in it different from the unit such that $\;a=a^{-1}\;$ . These two pieces are exactly the same question. – 2017-01-31

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we have $a^2=e$ then $a=a^{-1}$, let $A=\{g\in G; g\ne g^{-1} \}$ then $e\notin A$ ,if $g\in A \Rightarrow g^{-1}\in A$, then we have $|A|$ even number and $|A\cup\{e\}|$ odd number, therefore there exists $a\in G$ and $a\notin A \cup \{e \}$ where $a=a^{-1}$ i.e $a^2=e$