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.


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

  • $\begingroup$ This helped a lot. Thank you. $\endgroup$ – Stiven G Jun 29 at 14:37

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).


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