I'm stuck at this question. Can someone please help me?

Prove that if a group contains exactly one element of order 2, then that element is in the center of the group.

Let $x$ be the element of $G$ which has order 2. Let $y$ be an arbitrary element of $G$. We have to prove that $x \cdot y = y \cdot x$.

Since $x$ has order $2$, \begin{equation} x^2 = e \end{equation} That is, \begin{equation} x^{-1}=x \end{equation}

I don't really know how to proceed. I've tried a number of things, but none of them seem to work.

  • 1
    You also need to use the fact that no other element is of order 2 (so no other non-identity element is self-inverse). – M. Vinay Jun 5 '14 at 5:48
up vote 28 down vote accepted

Consider the element $z =y^{-1}xy$, we have: $z^2 = (y^{-1}xy)^2 = (y^{-1}xy)(y^{-1}xy) = e$. So: $z = x$, and $y^{-1}xy = x$. So: $xy = yx$. So: $x$ is in the center of $G$.

  • 1
    Does this proof require/use the fact that $x$ is an unique such element? – Agnishom Chattopadhyay Sep 10 '16 at 15:20
  • 3
    How do you know that $z=x$? couldn't $z=e$? you should add that if $z=e$ then $y=xy$, which means that $x=e$, but we know that $x \neq e$ since $x$ has order $2$. – Borat Sagdiyev Sep 25 '16 at 19:30
  • @AgnishomChattopadhyay Yes, because that's how you know that $z=x$ as opposed to some other element of order $2$ – Borat Sagdiyev Sep 25 '16 at 19:34
  • $z^2 = (y^{-1}xy)(y^{-1}xy)$ simplifies to $z^2 = x^2$, but how do you go from that to $z = x$? It could be that $z$ and $x$ are two elements that both have order 2, but they are different. – Goose Bumper Nov 3 '16 at 0:42
  • You have $z^2 =x^2 = e$, and $x$ is the only element of the group $G$ which is its own inverse, i.e $x^2 = e$, and since $z^2 = e \implies z = x$. – DeepSea Nov 3 '16 at 0:59

More generally, if a group$~G$ contains exactly one element$~x$ having any given property that can be expressed in the language of group theory (in particular without mentioning any specific element of$~G$, other than the identity element$~e$), then $x$ is in the center of$~G$. Namely, any automorphism of$~G$ must send $x$ to an element with the same property, which means it has to fix$~x$. In particular this is the case for inner automorphisms (conjugation by some element of$~G$), and this implies that $x$ is in the centre of$~G$.

  • Isn't that just relations of the form $x^n=e$, I can't think of anything else – Arun Kumar Jun 5 '14 at 13:40
  • @ArunKumar: No, it can be much more general, since there can be quantified variables that range over the group. The only thing not allowed is constants explicitly designating specific elements (using which one could of course characterise any one element). – Marc van Leeuwen Jun 5 '14 at 14:33

Hint: $$ y y^{-1} = e \implies yxxy^{-1} = e \implies yxy^{-1} yxy^{-1} = e \implies \left( yxy^{-1} \right)^2 = e \overset{*}{\implies} yxy^{-1} = x $$

Now the real question is why do we have the implication denoted by $\overset{*}{\implies}$?

Every element of a conjugacy class has the same order. Since there is only one element of order 2 that element forms a singleton conjugacy class. An element has a singleton conjugacy class iff it is in the center.

These are basic observations once you get to the class equation.

If a group contain exacly one element of order 2.then it has exactly one sub group order 2.so it must be normal .so for any Y belongs to G yxy^-1 should be in x .and it shoud be x so from that we say that X must be in centre of G

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