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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.

Anik Bhowmick
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user154185
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    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 05 '14 at 05:48

7 Answers7

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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 $G$.

DeepSea
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    Does this proof require/use the fact that $x$ is an unique such element? – Agnishom Chattopadhyay Sep 10 '16 at 15:20
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    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
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    @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 03 '16 at 00:42
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    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 03 '16 at 00:59
  • Oh I see. I'm starting by trying to prove that $x$ is the only element with order 2. In that case I can't make the assumption that $x$ is the only element in $G$ which is its own inverse. – Goose Bumper Nov 03 '16 at 06:03
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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$~G$.

Marc van Leeuwen
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  • Isn't that just relations of the form $x^n=e$, I can't think of anything else – Arun Kumar Jun 05 '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 05 '14 at 14:33
  • Hm. I've for a long time been trying to formalize to myself what exactly is meant by "natural" in "natural isomorphism" and "no distinguished basis", and your post gave me an intuitive idea: Because in the same way that "has an element named '$a$'" or "is named '$r$'" aren't really properties of groups or their elements (since they depend on iso type) whereas "has infinite order" and "is central" are, there is no way to single out or get at the elements $(0,\dots,0,1,0,\dots,0)$ in euclidean space no matter how "natural" or unbiased they look to me, but I can still talk about "rotations" of – Vandermonde May 03 '20 at 18:31
  • a vector space in that I can distinguish/recognize linear automorphisms of finite order (OK, so that's more like how a rational rotation should be. Maybe it was a bad idea trying to do this without a norm.). Thank you for your answer! It also reminds me of a transfer principle of sorts. – Vandermonde May 03 '20 at 18:34
  • This is an interesting answer. Where can I read more about this theorem? – Shaun Jan 21 '23 at 05:42
  • @Shaun There is no theorem, just the fact that (pure) statements of group theory cannot change truth value between isomorphic groups. If something is true in a group then transport by an isomorphism leads to a true statement at the target (which might be the same group). – Marc van Leeuwen Jan 25 '23 at 06:18
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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.

Eric Towers
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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 $\overset{*}{\implies}$?

DanZimm
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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

sanjoy
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For any elements $a$ and $x$ in a group $|xax^{-1}|=|a|$, since there is only one element with the same order as $a$ we have $xax^{-1}=a$ which implies that $xa=ax$. Hence $a$ lies in the centre

Daniel Aricatt
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For that matter: if $G$ is a group with exactly one element of order $n \gt 1$, then $n=2$ (and this element must be central as shown above). If the order of such an element $g$ is $n$, then in $\langle g \rangle$, there are $\varphi(n)$ elements of the same order. This forces $\varphi(n)=1$, whence $n=2$, since $n \gt 1$.

Nicky Hekster
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