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Show that for any group $G$ with an even numbers of elements, there is $x\in G$ and $x\neq e$ so that $x^2=e$

So we have that $|G|=2n$, $(n\in \mathbb{N})$ and we need to show that there is an element such that $o(x)=2$

How should I approach this?

And in general what is the connection between the order of a group and the order of its elements?

gbox
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  • Suppose for each $x \in G$ with $x \neq e$, we have $x^{2} \neq e$. Then the following is true:

    There exists $x \in G$ with $x \neq e$, since otherwise, $G$ would only have one element (the element $e$), implying $G$ has an odd number of elements.

    So let's name this element $x_{1}$. Then $x_{1} \in G$, and so we know by the axioms of a group that there exists an element $x_{2} \in G$ such that $x_{1}x_{2} = x_{2}x_{1} = e$. We call $x_{2}$ the inverse of $x_{1}$. We also know $x_{2} \neq e$ (why?) and $x_{1} \neq x_{2}$ (why?).

     – layman Dec 19 '16 at 16:35
  • (cont.) Now, we know there exists $x_{3} \in G$ such that $x_{3} \neq x_{1}$ and $x_{3} \neq x_{2}$ and $x_{3} \neq e$, since otherwise, the group $G$ would only have three elements (i.e., $G$ = {$x_{1}, x_{2}, e$ }), an odd number.

    But $x_{3}$ must have an inverse, too, and this inverse cannot be equal to $x_{1}$ or $x_{2}$, or even $x_{3}$ itself, nor $e$ (why?). So we have a distinct element $x_{4}$ in $G$ such that $x_{3}x_{4} = x_{4}x_{3} = e$. Now, if we stop here, we will have $G = {e,x_{1},x_{2},x_{3},x_{4}}$, which is a $5$-element group.

     – layman Dec 19 '16 at 16:35
  • (cont.) But our $G$ has an even number of elements. So there must exist a distinct $x_{6}$ such that...

    So, proceeding inductively as above, you can show for each $n \in \Bbb N$, $|G| > n$, implying $|G|$ is not finite, which contradicts the original assumption of $G$ being a finite group. Thus, there must exist an element $x \in G$ so that $x^{2} = e$.

     – layman Dec 19 '16 at 16:35

3 Answers3

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Suppose there is only $e$ solving $x^2=e$. Then we have for every $x\in G\setminus \{e\}$ that $x^2\neq e$ and so $x\neq x^{-1}$. Pairing every element with its inverse and counting we get an odd number of elements, a contradiction.So there must be an element $x\in G\setminus\{e\}$ with $x^{-1}=x$ and so $x^2=e$.

For every element $x\in G$ the (element-)order of $x$ is the (group-)order of the subgroup generated by $x$ in $G$

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Hint: Count the number of elements of order $1$, $2$, and $3$ or above separately. Note that each element of order $3$ or higher can be paired with its inverse.

Nitin
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By Sylow, you have a subgroup $H$ of order $2^m$, $x\in H$, $x$ $\neq e$, the order of $x$ is $2^l$, $x^{2^{l-1}}$ has order 2.

Tsemo Aristide
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