Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question

$S_3\times\Bbb Z_4$ 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 $8$.

share|cite|improve this answer
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 $Z_4$.

share|cite|improve this answer
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 $S_2$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.