Let $G=\langle x, y\mid x^7=y^3= e, yxy^{−1}=e\rangle$. Find $|G|$. Find a group that is isomorphic to $G$. Explicitly state the isomorphism

Let $G = \langle x, y \mid x^{7} = y^{3} = e, yxy^{−1} = e\rangle$. What is $|G|$? Find a familiar group that is isomorphic to $G$. Explicitly state the isomorphism.

I am trying to get a handle on understanding group presentations. At first glance, the group has at most 21 elements, however, I noticed: $$yxy^{−1} = e$$

implies that

$$yx = y \rightarrow yx = ye$$

Therefore, $x = e$ by the cancellation law. From my understanding, we're left with two possibilities, $|G| = 3$ or $|G| = 1$. The latter being the case that $y = e$ as well. I don't know how to narrow down answer.

If it is the case that $y = e$ then $y = y^{-1}$, but $y^{3} = e$ implies that $y^{2} = y^{-1}$...I thought that this would be enough to conclude that $|G| = 3$, but that last statement holds even when $y = e$.

Any guidance anyone can offer will be greatly appreciated. Is there a formal approach I'm completely missing to tackling questions like these?

From $yxy^{-1}=e$, we do indeed get

$$\begin{align} x&=exe\\ &=(y^{-1}y)x(y^{-1}y)\\ &=y^{-1}(yxy^{-1})y\\ &=y^{-1}ey\\ &=e. \end{align}$$

Therefore, the relation $x^7=e$ becomes trivial, along with $yxy^{−1}=e$. We are left with

$$\begin{align} G&\cong\langle y\mid y^3\rangle\\ &\cong \Bbb Z_3, \end{align}$$

where the final isomorphism is given by

$$\begin{align} \varphi: G&\to \Bbb Z_3,\\ x&\mapsto [0]_3,\\ y&\mapsto [1]_3, \end{align}$$

where

$$[a]_3=\{ b\in\Bbb Z: 3\mid a-b\}.$$

The order of $G$ is clearly, then, three, since isomorphisms preserve order.