# The normal subgroup of $\langle a, b\mid a^{22}, b^{15}, ab=ba^3\rangle$ generated by $a^2$.

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 :)

• 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. – David Cohen Aug 10 '17 at 22:41
• Anyway, your mistake is when you compute that $b^m a^{2k} b^{-m}$ is $a^{2k-3m}$. – David Cohen Aug 10 '17 at 22:47

## 1 Answer

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!