# Find a subgroup of $S_3\times\mathbb{Z}_4$ of size 8

8 should be a lcm of the orders of the components of the tuple. But 6 and 4 don't have any factors that have an lcm of 8. So how is this possible?

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$S_3\times\Bbb Z_4$ has order $6\cdot 4=24$, not $12=\operatorname{lcm}(6,4)$, and $8$ certainly divides $24$. Remember, you’re looking for a subgroup of order $8$, not an element of order $8$.

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GAP shows that there are 3 subgroups of order 8 with 3 generators. – Babak S. Oct 11 '12 at 8:43

Take the element $(s,0)$ which has order $2$ and the element $(1,1)$ which has order $4$. These two elements generate a group of order $8$. When I write $(1,1)$ the first $1$ is the multiplicative identity of $S_3$, while the second $1$ is the class of $1$ in $Z_4$.

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Since there are 3 elements of order 2 in $S_3$ (the three transpositions) there should be 3 such subgroups. – PAD Oct 11 '12 at 8:47

$\Bbb Z_4$ has order 4. If you can find a subgroup of $S_3$ with 2 elements—call it $T_2$—then $T_2\times \Bbb Z_4$ will be a subgroup of $S_3\times\Bbb Z_4$ of order $2\times 4 = 8$.

But subgroups of order 2 of $S_3$ are easy to find, because $S_3$ contains $S_2$.

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