Let $n$ be a multiple of $3$, i.e. $n = 3k, k \geq 1$ and consider the group
$$X_{2n} := \langle x,y|x^n = y^2 = 1, xy = yx^2 \rangle $$
Show that $|X_{2n}| = 6$.
(Source problem: dummit and foote, abstract algebra, p28 exercise 17)
Attempt: I managed to show that $|X_{2n}|\leq 6$ because I already have derived that $x^3 = 1$ and the commutation relation tells us that we can write every element in the form $y^i x^j$ with $0 \leq i \leq 1, 0 \leq j \leq 2$ so we have $6 = 2.3$ elements at most.
Now, $D_6$ is a group that satisfies the same relations as we have in $X_{2n}$, thus we must also have $|X_{2n}| \geq 6$. I'm unsure how I can formally justify the last step. Maybe make some embedding from the dihedral group in the given group?
Any input will be appreciated!
Note: I know what a homomorphism is, but at this point the book didn't introduce it (formally or unformally), nor did the book introduce presentations and free groups formally.
Best Answer
A group with given presentation is the largest group on the given generators satisfying the given relations.
Since you have proved $\vert X_{2n}\vert\le6$, we just need a group of order six with the same presentation.
$D_6$ is such a group: because once we know $x^3=e$, then $xy=yx^2\iff xy=yx^{-1}$.