0
$\begingroup$

Let $G=S_3 \times \mathbb{Z}_4$ be a group and $H= \langle(1 2 3)\rangle \times \langle2\rangle$ and $K=\langle(1)\rangle \times \langle2\rangle$ be subgroups.
Find the order of $G$, $H$ and $K$.
Find the left cosets of $H$ in $G$ AND $K$ in $H$.

$\endgroup$
3
  • $\begingroup$ Where are you stuck? $\endgroup$ – lhf Dec 5 '13 at 2:13
  • $\begingroup$ i don't know how to start $\endgroup$ – mariert Dec 5 '13 at 2:17
  • 1
    $\begingroup$ I know that S_3= (1)(12)(13)(23)(123)(321) and Z_4= {0, 1,2,3 } $\endgroup$ – mariert Dec 5 '13 at 2:19
1
$\begingroup$

The direct product of two groups $H$ and $K$ has the underlying set $$\{(h,k):h \in H \text{ and } k \in K\}$$ which has order $|H \times K|=|H|\,|K|$.

In this case, $S_3$ is the symmetric group on $3$ elements, so has order $3!=6$, and $\mathbb{Z}_4$ is the cyclic group of order $4$.


The notation $\langle h \rangle$ around a group element $h$ describes the subgroup generated by $h$. In this case, we have $\langle (123) \rangle=\{\mathrm{id},(123),(123)^2\}$ and $\langle \mathrm{id} \rangle=\{\mathrm{id}\}$, subgroups of $S_3$, and $\langle 2 \rangle=\{0,2\}$, a subgroup of $\mathbb{Z}_4$.


Cosets of subgroups are their "translates". Formally, (and using multiplicative notation) if $H$ is a subgroup of $G$, then the cosets are $gH=\{gh:h \in H\}$ for $g \in G$.

So, for example, the subgroup $\langle 2 \rangle$ of $\mathbb{Z}_4$ has the cosets $0+\langle 2 \rangle=\{0,2\}$ and $1+\langle 2 \rangle=\{1,3\}$ (we use additive notation for this group).


Your task is to apply these to the specific groups in the question. It's probably a difficult task if you haven't played around with these properties beforehand to get a feel for how they work.

$\endgroup$

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.