Find a group $G$ and elements $x,y,z ∈ G$ so that $|x|= 5$, $|y| = |z| = 7$, and $|xy| = 35$ but $|xz| ≠ 35$. I am doing an introductory group theory course, however this is one of the more advanced questions I have faced. This is my first question on here, so apologies if it isn't asked as well as it should be!

Question:

Find a group $G$ and elements $x,y,z ∈ G$ so that $|x|= 5$, $|y| = |z| = 7$, and $|xy| = 35$ but $|xz| ≠ 35$. 

My attempt:

If we take $G$ as finite, $|G|$ must be divided by both $5$ and $7$, by Lagrange's Theorem. 
This would mean the order of $G$ must some multiple of $\operatorname{lcm}(5,7)$, which is $35$.
In this case I can't see how $|xz|$ could be anything other than $35$, as I think it can't be of order $1$, $5$, or $7$, the other divisors of $35$. 

By this I feel I should be looking for an infinite Group, however I am very unfamiliar with these and haven't been able to come up with any examples.

If anybody could tell me is my above reasoning correct, and point me in the right direction I would really appreciate it!
Orders of products of group elements can be quite unintuitive. For example, a lot of simple groups can be generated by two elements with order $2$ and $3$.

If you are familiar with cycle notation in $S_n$, consider $x=(1,2,3,4,5)$, $y=(6,7,8,9,10,11,12)$ and $z=(1,2,3,4,5,6,7)$ (I had to add commas because you get numbers with two digits).

Then $xz=(1,3,5,6,7,2,4)$, which has still order $7$.

The fact that $7$ divides $35$ is... a coincidence. You can do "worse" things! For example if you take $z=(5,6,7,8,9,10,11)$ you get $xz=(1,2,3,4,5,6,7,8,9,10,11)$ which has order $11$.
Hint: If $x$ and $y$ commute and $\gcd(|x|, |y|)=1$, then $xy$ has order $|x||y|$.