order of group generated by two element with some relation. The group defined by generators $a,b$ and relations $a^{8}=b^{2}a^{4}=ab^{-1}ab=e$ has order at most 16. How to prove that? I have no idea.
 A: Try writing out the elements: they're arbitrary strings of $a$s and $b$s, perhaps with exponents. Then notice that the exponent on $a$ can be between 0 and 7 (every $a^8$ can be replaced with $e$). 
Then show that any any sequence like $aba^2b^3$ con be converted to one of the form $b^j a^i $. (Hint: that last relation will help, if you rewrite it as $ab = b a^{-1}$). Then argue that $b$ never needs to appear with a power greater than $2$. Pretty soon you've got at most 16 elements, as $i$ ranged from 0 to 8, and $j$ ranges from 0 to 1. 
(I"m not sure this exact process will work, but something very like it will, and it's how I generally approach these things.)
Let's look at "trying to move the $b$s to the front." If you have a sequence that ends with a $\ldots ab^k$, you can change it so that it ends with $b^{k-1}$ by
\begin{align}
a b^k &= a b b^{k-1} \\
      &= (b a^{-1}) b^{k-1}
\end{align} 
using that fourth relation. 
What if it ends with $a^{-1} b^k$? Treat that as $a^7 b^k = a^6 a b^k$, and you've reduced to the previous case. By induction, you can conclude that any sequences of $a$s and $b$ that ends with a nontrivial power of $b$ can be converted to one that ends with $a$ instead, and has the same number of $b$s. With a bit more work, you can conclude that all the $b$s can be moved to the front. 
@AWertheim asks how you'd move the $b$ to the front in $aba^2$. The idea is to ignore the trailing $a$, so you've got a simpler problem. Here's how:
\begin{align}
a b a^2 &= (a b) a^2 \\
      &= (b a^{-1}) a^2\\
&= ba
\end{align} 
More interesting, perhaps, is $a^2b$ or $a^3 b$. Let's do the first:
\begin{align}
a^2 b  &= a (a b) \\
      &= a (b a^{-1})\\
      &= (a b) a^{-1})\\
      &= (b a^{-1}) a^{-1})\\
    &= ba^{-2}
\end{align} 
After this, it's pretty clear that $a^k b = b a^{-k}$, and with that, it's easy to move the $b$s to the front: convert every exponent on every $a$ to be positive, by adding $8$ enough times; then use the rule above to move the $b$s leftwards. 
A: The above answer properly proves that every element of the group can be written as $b^ka^j$ with $0 \leq j < 8$. JH forgot to show that you can restrict $k$ to zero or 1.  
The completion of the proof is not hard:  If the string of $b^k$ is longer than one (that is, if $k>1$), re-write the last two using $b^2 = (a^{-1})^4$ and absorb those $a^{-1}$'s into the string of $a$'s. You now have two fewer $b$'s at the start, so you can always get down to $k=1$ or $k=0$.
