Inverse of elements in a group If $x,y,z$  are elements of a  group such  that $xyz=1,$ then which  of  the  following are true?


*

*$yzx=1$

*$yxz=1$

*$zxy=1$

*$zyx=1$


I have found options 1  and  3  to  be  correct, but how  to  prove  that options 2 and 4 are wrong (that is,  what  the  given  answers  say)?
 A: 1 and 3 are true, which follows from a little more general result: if $x_1 x_2 \dots x_n = 1$ and $\pi$ is a cyclic permutation, then $x_{\pi(1)}x_{\pi(2)}\dots x_{\pi(n)} = 1$.
To prove that 2 and 4 are false, it is sufficient to provide counterexamples.
2) Let $z = 1$, then $xyz$ = $xy$ and $yxz$ = $yx$. Take $x$ and $y$ to be  non-commuting elements of some group. Then $xy \neq yx$ and $xyz \neq yxz$.
4) Let $y = 1$, then $xyz$ = $xz$ and $zyx$ = $zx$. Take $x$ and $z$ to be  non-commuting elements of some group. Then $xz \neq zx$ and $xyz \neq zyx$.
For an example of non-commuting pairs of elements, you can use $\mathbb{S}_3$ (group of permutations on 3-element set) and take transpositions $(12)$ and $(13)$, which do not commute: $(12)(13) = (231)$ and $(13)(12) = (312)$.
Note that all 4 statements hold in an abelian group.
A: You have
$$
xyzz^{-1}=1z^{-1}
$$
so
$$
xy=z^{-1}
$$
hence
$$
zxy=zz^{-1}=1
$$
Similarly, you can show that $yzx=1$. So you're right in saying that 1 and 3 hold *without any other assumption on the group.
Since 2 and 4 obviously hold when the group is abelian, you can find a counterexample only in a non abelian group, the simplest one is $S_3$.
Consider $x=(12)$, $y=(123)$, $z=(23)$ and do the computations.
A: $1$ is always true. We are told $x$ is the inverse of $yz$ and so $yzx=1$. 
analogously $3$ is always true. We are told $xy$ is the inverse of $z$ and so $zxy=1$

The others are not true, for $2$ notice if $xy\neq yx$ then they can't  have the same inverse. (examples of $xy\neq yx$ exist in any non-abelian group)
$4$ is slightly trickier, it combines the ideas of $1$ and $2$.
You can solve it as follows: If $zyx=1$ then $yxz=1$ (since $z$ and $yx$ would be inverses). Notice this is problem $2$.
