This is Exercise 12.2(a) of Roman's "Fundamentals of Group Theory: An Advanced Approach". According to this search, it is new to MSE.
The Details:
Roman defines, on page 350 ibid., the dicyclic group of order $4n, n>1$, using the presentation
$${\rm Dic}_n\cong \langle x,y\mid x^{2n}, y^2=x^n, yx=x^{2n-1}y=x^{-1}y\rangle.\tag{$\mathcal{P}$}$$
The Question:
Why does the dicyclic group have exactly one involution?
Thoughts:
Since the order of the dicyclic group is $4n$, two divides its order, so by Cauchy's Theorem, an involution exists; but this is nuking the problem, since $y^2=x^n$ is an involution by the relation $x^{2n}=1$ in $(\mathcal{P})$.
Suppose, then, for a contradiction, that $z^2=1$ for some $z\in{\rm Dic}_n$ with $z\neq x^n$. Now what do I do?
Experience tells me that we can write each element of ${\rm Dic}_n$ in the form $x^ay^b$ for some $a\in\overline{0,2n-1},$ $b\in\{0,1\}$. This follows from the orders of $x$ and of $y$, together with the last relation in $(\mathcal{P})$. I'm fairly sure this will help.
An Example:
Let $n=2$. Then ${\rm Dic}_2$ is isomorphic to the quaternion group. Clearly $-1$ is its only involution.
My Background:
I work with presentations a lot, so this seems like a problem I should be able to solve myself with a bit of a hint. I don't have much experience with dicyclic groups.
Further Context:
This question is from a rather introductory chapter titled "Free Groups and Presentations", so the tools available aren't as sophisticated as in, say, Magnus et al.'s "Combinatorial Group Theory: [. . .]".
Please help :)
Well, it is clear that one involution is $x^n$, and this is the only power of $x$ that is an involution. Every other element can be written in the form $x^iy$ with $0\leq i\lt 2n$, using the given relations. Note that any even power of $y$ can be written as a power of $x$, since $(y^{2k}) = (y^2)^k = (x^n)^k = x^{kn}$, and any odd power of $y$ can be written as a power of $x$ times $y$, since $y^{2k+1} = y^{2k}y = x^{kn}y$. That’s why every element can be written as $x^iy^j$ with $0\leq i\lt 2n$, $0\leq j\lt 2$. Uniqueness takes a bit more work, but isn’t terribly hard.
We have $$(x^iy)^2 = x^iyx^iy = x^ix^{-i}y^2 = x^n.$$ So this is trivial if and only if $x^n=e$, which cannot happen since $x$ has order $2n$. Thus, none of these elements are involutions, and so the only involution is $x^n$, which we already found.