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 $H$.

  • $\begingroup$ 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

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.


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