What do I have that to verify to prove that G has order 16.? The group defined by $G=\langle a,b:a^{8}=b^{2}a^{4}=ab^{-1}ab=e\rangle$ has order at most 16.
What do I have that to verify to prove that G has order 16.?
Any suggestions to verify that has order 16. Thanks.
 A: The last equation implies that $A:=\langle a\rangle$ is normal in $G$. It's easy to see from the presentation that $|G/A|\leq 2$ and $|A|\leq 8$ thus $|G|\leq 16$.
It's not hard to check that $G$ does have order 16. In fact, it is a generalised quaternion group:
https://en.wikipedia.org/wiki/Quaternion_group#Generalized_quaternion_group
https://en.wikipedia.org/wiki/Dicyclic_group
http://groupprops.subwiki.org/wiki/Generalized_quaternion_group:Q16
A: Hint: Write out its elements.  You know that every element is a product of $a$'s and $b$'s.
Since $a^8=1$, start with $\{e,a,\cdots,a^7\}$.
Next, multiply by $b$ to get $\{b,ab,\cdots,a^7b\}$.
Take these sets and multiply by $a$ or $b$, at every step, you'll either get a set that you've already seen or something new.  When you run out of new sets, you're done.
Also, observe that the last equation, $ab^{-1}ab=e$ can be rewritten as $ab=ba^7$.  This makes the second set above $\{b,ba^7,ba^6,ba^5,\cdots,ba\}$.  It may also help to prove that the order of $b$ is finite (this comes from $b^2a^4=e$, so $b^2=a^4$ and the order of $a$ is finite).
