# Show that S6 has at least 60 subgroups of order 4

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.

• There are far more than 6 cyclic subgroups of order 4. Consider the subgroup generated by $(1425)$ for example. – MJD Mar 8 '19 at 3:37
• Thank you very much for the hint, that was my fault. – Rico Mar 8 '19 at 3:39
• 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. – Rico Mar 8 '19 at 3:41
• @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. – Robert Shore Mar 8 '19 at 4:13
• @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 $45$. – Robert Shore 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.

• Also subgroups like $\left<(12)(34),(56)\right>$, $\left<(12)(34),(13)(24)(56)\right>$ and others.... – Angina Seng Mar 8 '19 at 6:00
• 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. – Teresa Lisbon Mar 8 '19 at 6:04
• 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. – verret Mar 14 '19 at 22:21
• I see. Thank you for the comment., I could not find the exact number online – Teresa Lisbon Mar 15 '19 at 2:01