Every group of order 120 has a subgroup of index 3 or 5

I have been reviewing basic algebra over this summer as preparation for my PhD quals, and took this opportunity to solve some problems from Isaacs’s Finite Group Theory. All of the exercises in section 1C, which deals with the Sylow theorems, were reasonably easy, except for 1C4, which asks us to prove the statement given in the title. I felt quite good after actually solving this problem, and so I thought it would be apt to post my solution as the inaugural post for this blog.

Acknowledgement: I had basically covered some trivial cases one night, and called up my good friend Saraswata to cross-check if I had made any silly errors till then. I went to bed afterwards without making any serious progress, but somehow when I sat down with the problem again the next morning, I was able to essentially bypass the case where I had gotten stuck; thanks to Saraswata for listening to me ramble at 2 am

Notation:

For a finite group \(G\) and a prime \(p\), we denote the set of \(p-\)Sylow subgroups of \(G\) by \(\mathrm{Syl}_p(G)\) and also set \(n_p(G) = \# \mathrm{Syl}_p(G)\). We omit the mention of \(G\) if it’s clear from context. We will use the following generalization of the Sylow subgroup counting theorem, proved in Isaacs as Theorem 1.16:

Suppose that \(G\) is a finite group such that \(n_p(G) > 1\). Let \(S\) and \(T\) be Sylow-\(p\) subgroups of \(G\) such that \(\vert S \cap T \vert\) is maximal. Then, \(n_p(G) \equiv 1 \pmod{[S:S \cap T]}\).

A nice example application of this result is discussed in the book right after the statement, and I encourage everyone to take a look at that. Essentially, we use the result to get stronger bounds on the index \([S:S \cap T]\), which in turn enables us to give arguments about normality of Sylow subgroups and so on. Let’s now dive into the actual solution.

The solution:

Let \(G\) be a group of order 120. Note that \(120 = 2^3 \cdot 3 \cdot 5\). By the Sylow theorems, we thus have

\(n_2 \equiv 1 \mod 2\) and \(n_2 \mid 15\), so that \(n_2 \in \{1,3,5,15\}\).
\(n_3 \equiv 1 \mod 3\) and \(n_3 \mid 40\), so that \(n_3 \in \{1,4,10,40\}\).
\(n_5 \equiv 1 \mod 5\) and \(n_5 \mid 24\), so that \(n_5 \in \{1,6\}\).

We will deal with the 4 possible values of \(n_2\) separately. Suppose first that \(n_2=1\), and let \(N \vartriangleleft G\) be the Sylow-2 subgroup. Then \(\vert G/N \vert = 15\). But every group of order \(pq\), where \(p < q\) are distinct primes such that \(q \not\equiv 1 \mod p\), is known to be cyclic, and so we get \(G/N \simeq \mathbb{Z}/ 15\mathbb{Z}\). We also know that this has a subgroup \(K/N\) of order 3, and hence, we get a subgroup \(K \leq G\) of index 5.

We assume \(n_2 > 1\) henceforth. If either one of \(n_3,n_5\) is 1, we pick this \(3-\) or \(5-\)Sylow subgroup \(P\) and consider the product \(PQ\), where \(Q \in \mathrm{Syl}_2\): this is a subgroup because of the normality of \(P\), and has order either 24 or 40 (because \(P \cap Q = \{e\}\)), and thus index either 5 or 3. In either case, we are done; we are now left the cases where \(n_5=6\) and \(n_3>1\).

Consider a non-maximal \(P \in \mathrm{Syl}_2\), i.e, there is a subgroup \(H\) such that \(P < H < G\). But note that \(\vert P \vert = 8\) must divide \(\vert H \vert\), which gives us the only possibility is that \(\vert H \vert\) is either 24 or 40, and so we are done in this case as well. Therefore, we can assume all \(2-\)Sylow subgroups are maximal, without any loss of generality.

By maximality, \(N_G(S)\) is either \(S\) or \(G\) for all \(S \in \mathrm{Syl}_2\). As \(n_2 > 1\), we in fact get \(N_G(S) = S\) for all \(2-\)Sylow subgroups \(S\). But, the Sylow theorems then give us

\[n_2 = [G: N_G(S)] = [G:S] = 15\]

Let’s take stock of where we are: we have shown the existence of the required subgroup in all but 3 cases, namely when \(n_2 = 15, n_5 = 6\) and \(n_3>1\). The time is now ripe to use the result mentioned previously. We pick \(S,T \in \mathrm{Syl}_2\) such that \(D := S \cap T\) is of maximal order. The result then gives us \(n_2 = 15 \equiv 1 \mod [S:D]\). Note that \([S:D] = \frac{8}{\vert D \vert}\). As \(D\) must be properly contained in both \(S\) and \(T\), it has order less than 8. This leads to the only possibility that \([S:D] = 2\), i.e, \(\vert D \vert = 4\).

As \(D < S\) is a subgroup of index 2, it is normal. By the symmetry of our arguments in \(S\) and \(T\) so far, we actually have \(D\) is normal in both of them, i.e, \(S,T \subseteq N_G(D)\). Note that equality here would imply \(S = T\), which then means \(\vert D \vert = 8\), a contradiction! Hence, by the maximality of \(2-\)Sylow subgroups, \(N_G(D) = G\). Thus, we get \(D \vartriangleleft G\).

We now consider the group \(\bar{D} = G/D\), which has order \(30 = 2 \cdot 3 \cdot 5\). Suppose we have managed to find a subgroup \(H/D \leq \bar D\) of order 6 or 10: in either case, we get a subgroup \(H \leq G\) whose index is

\[[G:H] = \frac{[G:D]}{[H:D]} \in \{3,5\}\]

Therefore, should we manage to prove the existence of such a subgroup of \(\bar D\), we would finish the proof. Applying the Sylow theorems to \(\bar D\), we get \(n_3(\bar D) \in \{1,10\}\) and \(n_5(\bar D) \in \{1,6\}\). Now (repeating the argument given in the beginning) if either one of \(n_3, n_5\) is 1, we can again take the product of this normal Sylow subgroup with any \(2-\)Sylow subgroup of \(\bar D\) to get a subgroup of order 6 or 10. Hence, it remains to show that at least one of \(n_3(\bar D)\) and \(n_5(\bar D)\) must be 1.

We argue this last bit by contradiction: suppose \(n_3(\bar D) = 10\) and \(n_5(\bar D) = 6\). Because the exponent of both 3 and 5 in \(\vert \bar D \vert\) is 1, we can count the number of elements having orders 3 and 5:

For order 3: \(\# \{x \in \bar D \mid x^3 = \bar e, x \neq \bar e\} = n_3(\bar D) \cdot 2 = 20\)
For order 5: \(\# \{x \in \bar D \mid x^5 = \bar e, x \neq \bar e\} = n_5(\bar D) \cdot 5 = 24\)

Adding these two counts gives 44, which exceeds the order of \(\bar D\)! Hence, this situation cannot happen, and so we are done. \(\blacksquare\)

Do let me know if you find any mistakes or typos by commenting below, or even by emailing me. Hope you found this post useful!