Let $G=\langle a \rangle$ be a cyclic group and $H$ a subgroup of $G$. Show $(G/H,*)$ is a cyclic group with generator $aH$. Also find a group $K$ with normal subgroup $L$ such that $(K/L,*)$ is abelian, but $K$ is not abelian.
We can accept:
Lemma: Let $G$ be a group, $H$ a normal subgroup of $G$. Let $b \in G$, then $(bH)^n=b^nH$, $\forall n\in \Bbb{Z}$ .
So, I need help figuring this out. What I have so far is this:
Since $G$ is cyclic then it is abelian and we get the result that $H$ is a normal subgroup. From here let $G=\langle a \rangle=\{a,a^2,\cdots,a^m,e\}$. If we write out all the left cosets for $H$ in $G$ we have $gH=\{aH,a^2H,\cdots,a^mH,H\}$, $g\in G$. From here we can easily see that$$\langle aH \rangle=\{(aH)^1,(aH)^2,\cdots,(aH)^m,(H)^{m+1}\}=\{aH,a^2H,\cdots,a^mH,H\}.$$
Is this logically correct? And any hints on the second part of the question?
Best Answer
Your reasoning is pretty sound.
$G/H$ is the homomorphic image of $G$ via the canonical projection map $\pi$, with $\pi (a)=aH$. (In general, the homomorphic image of a cyclic group is cyclic, generated by the image of a generator.)
For part $2$, consider $K=S_3$ and $L=\{(123), (132),e\}$.
Or, try to find any nonabelian group that does not equal its commutator subgroup. (Like $S_n$ and $A_n$.) This is because the commutator is the smallest normal subgroup for which the quotient is abelian. (Try the quaternions, or $A_4$, for instance. The latter has commutator the Klein four group. The former also works. So these provide examples. Or investigate $\operatorname {GL}_n(\Bbb k)$, whose commutator is a proper subgroup, $\operatorname {SL}_n(\Bbb k)$, in most cases ($n\neq2$ or $\Bbb k$ not the field consisting of $2$ elements) See this.)
Or, an index $2$ subgroup is always normal (and, of course, the quotient abelian).