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Is there a standard name or qualifier for groups with exactly two self inverses, that is groups $(G,\star)$ with exactly two elements $x$ such that $x\star x=e$, where $e$ is the group's identity?

Can we call these $x$ "root of unity", or is that used only for fields? Is there some more suitable name for these $x$?

One of these two $x$ is $e$. Is there a standard name or notation for the other?

Does it help to find such vocabulary if we restrict to finite and/or commutative groups?

Example: the multiplicative subgroup of the field $\mathbb F_p$ with odd prime $p$ (sometime noted $\mathbb Z_p^*$), $e=1$. The other $x$ is $p-1$.

Shaun
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fgrieu
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    The equivalent definition is a group that has exactly one element of order $2$. I do not think that such groups have a specific name. – Peter Jan 20 '23 at 13:36
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    Another nice fact about such groups is, that then these two self-inverse elements must lie in the center - see here. – Dietrich Burde Jan 20 '23 at 13:42
  • @DietrichBurde It is however true for odd primes , as assumed in the example. – Peter Jan 20 '23 at 13:47
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    Also, all cyclic groups of even order have this property - see this post. So there are more examples. Indeed, every abelian group of order $2n$, with odd $n$ has also this property - see this post. – Dietrich Burde Jan 20 '23 at 13:51
  • If you didn't say "exactly two" but instead "with a specified non-identity involution" then maybe something like "involutive group" would be appropriate. But disallowing other involutions probably wouldn't be implied by a phrase like that. – Mark S. Jan 20 '23 at 15:14
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    Worth a look is Peter Cameron's blog: https://cameroncounts.wordpress.com/2011/06/22/groups-with-unique-involution/ – ancient mathematician Jan 20 '23 at 15:32
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    I think @DietrichBurde's comment essentially classifies these groups - they are the groups whose center's contain precisely one element of order $2$ and all non-central elements have order not $2$. For finite groups, this means that the center has the form $H\times\mathbb{Z}{2n}$ for $H$ of odd order and $n\in\mathbb{N}\setminus{0}$. There are plenty of groups which look like this, for example $K\times\mathbb{Z}{2n}$ where $K$ has odd order or has no elements of finite order or whatever. The quaternions form a non-trite example. – user1729 Jan 20 '23 at 15:37
  • Also $SL_2(R)$ for $R$ any domain with $1\neq -1$. – David A. Craven Jan 20 '23 at 16:17
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    So far the best candidates seem to be "Groups with unique element of order 2", and "Groups with unique involution" (the later by Peter Cameron). To my ears, "involution" takes the shortcut of identifying a group element and the function obtained by applying the binary operation with that group element, thus I'm more comfortable with the first. – fgrieu Jan 20 '23 at 16:33

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The following is too long to be a comment, so I have made it a CW post.

Note that $x^2=e$ if and only if $x=x^{-1}$.

Theorem: No group has exactly two elements of order two.

Proof: Suppose $G$ is a group with exactly two elements of order two. Call them $a$ and $b$. Consider $aba$. We have $$\begin{align} (aba)^2&=(aba)(aba)\\ &=(ab)a^2(ba)\\ &=(ab)(ba)\\ &=ab^2a\\ &=a^2\\ &=e. \end{align}$$

Thus $aba$ has order two (or $aba=e$, but that implies $b=a(aba)a=a^2=e$), so it must equal either $a$ or $b$.

Suppose the former. Then $aba=a$ implies $ab=e$, so $b^{-1}=a$. But the inverse of any order two element is the element itself.

Thus $aba=b$, which implies $ab=ba$. Now

$$\begin{align} (ab)^2&=(ab)(ab)\\ &=(ab)(ba)\\ &=ab^2a\\ &=a^2\\ &=e. \end{align}$$

Hence either $ab=a$ or $ab=b$. But the former implies $b=e$ has order one, whereas the latter implies $a=e$ has order one, which are contradictions.$\square$

Shaun
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