Show that $S_6$ has at least $60$ subgroups of order $4$. [Hint: Consider cyclic subgroups generated by a 4-cycle (such as $\langle(1234)\rangle$) or by the product of a 4-cycle and a disjoint transposition (such as $\langle(1234)(56))\rangle$; also look at noncyclic subgroups, such as ${(1), ( 12), (34), (12)(34)}$.]

For my work so far, I have found that the cyclic groups of order 4 are

$\langle(1234)\rangle, \langle(2345)\rangle,...,\langle(6123)\rangle$ which are $6$ of them

and for disjoint transposition of order 4 I have

$\langle(1234)(56)\rangle, .... ,\langle(6123)(45)\rangle$ which are also 6 of them

I not sure what the last hint is leading and I don't know how to proceed from here.

Side note: this is a question from the section related to Sylow's Theorems and Cauchy's Theorem. But I don't think they are helpful for this question. I feel it is just related to symmetric and alternating groups.

  • $\begingroup$ There are far more than 6 cyclic subgroups of order 4. Consider the subgroup generated by $(1425)$ for example. $\endgroup$
    – MJD
    Mar 8 '19 at 3:37
  • $\begingroup$ Thank you very much for the hint, that was my fault. $\endgroup$
    – Rico
    Mar 8 '19 at 3:39
  • $\begingroup$ May I ask the method of finding the numbers of these order 4 cyclic groups? I was considering fixing 1 and consider what number fits for the 3 other slots but I lose track of it in the end. $\endgroup$
    – Rico
    Mar 8 '19 at 3:41
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    $\begingroup$ @J.W.Tanner This overcounts the subgroups. $\langle(1234)\rangle = \langle(1432)\rangle$. But by my count, there are $30$ subgroups generated by $4$-cycles and another $30$ created by adding the disjoint transposition to a $4$-cycle. $\endgroup$ Mar 8 '19 at 4:13
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    $\begingroup$ @J.W.Tanner Yes, you're correct. I counted only the cyclic subgroups that don't stabilize $1$. When you add the others, there are $45$. $\endgroup$ Mar 8 '19 at 5:13

For merely the cyclic part, consider elements of the form $(abcd) \in S_4$. How many of them are there? Indeed, the answer seems to be just $6 \times 5 \times 4 \times 3 = 360$ by choosing $a,b,c,d$ in that order. However, noting from the comment that $(abcd) = (bcda) = (cdab) = (dabc)$ ( in the cycle representation, for example $(1234) = (2341)$ ) tells you that we must divide by $4$ to avoid repetition. This leads to $90$ cycles.

However, when we are looking at the generated cyclic group, indeed note that $(abcd)$ and $(adcb)$ generate the same group (because if $x$ is of order $4$ then $x^3 = x^{-1}$ is also of order $4$ and hence generates the same group. However, $x^2$ is not, so we will have to be careful there).

Also, if two elements generate the same subgroup of order $4$ then either they are the same or inverses. Therefore, the above analysis gives $45$ distinct groups of order $4$.

Now, for the other part, we consider groups generated by $(abcd)(ef)$ which would be of order $4$ but would not coincide with any of the previous groups since there is always an element which does not have any fixed points here.

Choosing $(abcd)(ef)$ happens again in $90$ ways, since if we choose $(abcd)$ then $(ef)$ gets fixed for us. Once again, going to the cyclic subgroup , we can pair $(abcd)(ef)$ with its inverse which is $(adcb)(ef)$ for each element of this kind, once again resulting in $45$ distinct groups of order $4$.

Totalling the above gives $90$ distinct cyclic groups of order $4$, more than what is required to answer the question.

Apart from this if we choose to look at groups generated by two transpositions $(ab) \neq (cd)$, which are distinct from the previous ones since every element here has order at most two , then $(ab)$ has $15$ choices and $(cd)$ has $6$ choices, after which you remove their order of picking to get $45$ further choices. This leads to at least $135$ subgroups of order $4$. You can try to check if some others are there, my guess would be no.

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    $\begingroup$ Also subgroups like $\left<(12)(34),(56)\right>$, $\left<(12)(34),(13)(24)(56)\right>$ and others.... $\endgroup$ Mar 8 '19 at 6:00
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    $\begingroup$ Yes, it seems that you all you need is $\langle x,y\rangle$ such that $x \neq y$ have order $2$ and $xy=yx$, which is the case for the examples you provide. $\endgroup$ Mar 8 '19 at 6:04
  • $\begingroup$ As you said, cyclic subgroups of order $4$ fall into two conjugacy classes of size $45$. For the ones isomorphic to the Klein Group, there are two classes of size $15$, and three classes of size $45$, so $255$ subgroups of order $4$ in total. $\endgroup$
    – verret
    Mar 14 '19 at 22:21
  • $\begingroup$ I see. Thank you for the comment., I could not find the exact number online $\endgroup$ Mar 15 '19 at 2:01

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