3
$\begingroup$

This is Exercise 2.1.5(b) of "Combinatorial Group Theory: Presentations of Groups in Terms of Generators and Relations," by Magnus et al.

The Question:

For each of the following groups $G$, let $H$ be the subgroup generated by the given elements. Show that $H$ is normal and find the order of $G/H$.

(b) $G=\langle a, b\mid a^{22}, b^{15}, ab=ba^3\rangle$; $a^2$.

My Attempt:

The elements of $G$ are of the form $a^nb^m$, where $n\in\{0,1,\dots ,21\}, m\in\{0,1, \dots ,14\}$, and the elements of $H$ are of the form $a^{2k}$ for $k\in\Bbb Z$, so we have $$\begin{align} a^nb^ma^{2k}(a^nb^m)^{-1}&=a^nb^ma^{2k}(a^nb^m)^{-1} \\ &=a^nb^ma^{2k}b^{-m}a^{-n} \\ &=a^nb^ma^{2k}b^{-m}a^{-n} \\ &=a^na^{2k-3m}a^{-n} \\ &=a^{2k-3m}, \end{align}$$

but I don't know what to do with the power $2k-3m$, which should be a multiple of $2$. Have I got it right so far? What do I do next?

As for the order of $G/H$, we have that $$\begin{align} G/H&\cong \langle a, b\mid a^2, b^{15}, ab=ba\rangle \\ &\cong C_2\times C_{15}, \end{align}$$ so that $\lvert G/H\rvert=30$.

Is that correct?

Please help :)

$\endgroup$
  • $\begingroup$ How did you get that elements of $G$ are of that form? To show that $\langle a^2\rangle$ is normal, you just need to check that it is preserved under conjugation by $b$ and $b^{-1}$ (since it is clearly preserved under conjugation by $a$ and $a^{-1}$). Since $b$ has finite order in $G$, it is enough to check that $b^{-1}a^2b\in\langle a^2\rangle$, which should be easy. $\endgroup$ – David Cohen Aug 10 '17 at 22:41
  • 1
    $\begingroup$ Anyway, your mistake is when you compute that $b^m a^{2k} b^{-m}$ is $a^{2k-3m}$. $\endgroup$ – David Cohen Aug 10 '17 at 22:47
3
$\begingroup$

Calculations are messy, and errors are easy to make (as you saw!). So lets do the question without calculations:

Clearly $b^{-1}ab\in\langle a\rangle$ by the relation $ab=ba^3$. On the other hand, $3$ is coprime to $22$ and so $\langle a\rangle=\langle a^3\rangle$. Therefore, $bab^{-1}\in\langle a\rangle$ by the relation $ab=ba^3$.* Hence, $\langle a\rangle$ is normal in $G$.

Finally, $\langle a^2\rangle$ is characteristic in $\langle a\rangle$ (why?) and hence is normal in $G$.

*I have no idea what $bab^{-1}$ actually is, but you do not need to calculate it to answer the question!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.