Which of the following statements are correct? T he order of the smallest possible non trivial group containing elements $x$ and y such that
$x^7=y^2=e$ and $yx=x^4y$ is
(A) $1\space\space$ (B) $2\space\space$ (C) $7\space\space$ (D) $14$ 
 A: You can just construct it with $\mathbb{Z}_2$ under addition (mod 2) and let $x=0$ and $y=1$.  If you add $0$ seven times you get $0$ and if you add $1$ twice, you get $0$ (mod $2$).  The last one is $$1 + 0 = 0+0+0+0+1 \pmod{2}$$
Not going to find any smaller non-trivial groups than one of order $2$.
Now if you knew the order of $x$ were $7$ and the order of $y$ were two, then things would be different.
A: This is how I thought about it.
First, it shouldn't be A, because the group is said to be non-trivial.
If it is B than the non-identity element has even order and you have $x^7=e$ which means that it only works if $x=e$. In this case the relations reduce to $e^7=y^2=e$ and $y=y$ and this is clearly possible. That answers the question.

However the wording of the question is somewhat loose. We can let $x=y=e$ so that the relations given are trivially satisfied, and because the question doesn't say that $x,y$ generate the group, we can introduce any other elements we like. Normally we wouldn't do that as it would just make the group bigger - but here we need to make it bigger to meet the "non-trivial" condition. So we can choose $x=y=e, z^2=e$ as an alternative group of order $2$.
A: If we read the question literally, we might assume that $e,x\ne e,y\ne e\in G$, so $|G|\ge2$. This immediately removes A and B, and we can argue as follows:
It cannot be A because this is the trivial identity group.
It cannot be B because we have $x=y$ and therefore the statement $x^7=y^2=e$ is inconsistent.
It cannot be C because we have $y=e$ and therefore $ex=x^4e$, so $x^3=e$ so $x=e$, which means the group has only $1$ element.
It cannot be D because we have $yx=y$, so $x=e$ but this means the group is order 2, but the claim for D is that the group is order $14$.
But this doesn't leave any correct answer, so we can assume either $x=e$ and/or $y=e$. This removes A and C as we have in both cases $x=y=e$, or $D$ as $x=e$.
If we assume $x=e$ both B and D are correct groups in principle but D claims the group is order $14$ whereas it is order $2$. B is a correct claim, and indeed the smallest non-trivial group.
