If $|G| = 120$ then $G$ has a subgroup of index $3$ or $5$ (or both)

This is exercise 1.C.4 in Isaacs, Finite Group Theory. I think I have a proof, but would like to verify the proof and also inquire whether it can be shortened or improved significantly.

Let $|G| = 120 = 2^3 \cdot 3 \cdot 5$. Show that $G$ has a subgroup of index $3$ or a subgroup of index $5$ (or both). Hint: Analyze separately the possibilities for $n_2(G)$.

My proof: By the Sylow theorems, $n_2(G)$ must be an odd integer which divides $15$, so $n_2(G) = 1$, $3$, $5$, or $15$.

Case 1: $n_2(G) = 1$. In this case, $G$ has a unique (hence normal) Sylow $2$-subgroup $S$ of order $8$ (index $15$). Then $G/S$ is a group of order $15$, so it has subgroups of order $3$ and $5$ (index $5$ and $3$, respectively). By the correspondence theorem, these subgroups of $G/S$ correspond to subgroups of $G$ which contain $S$ and which have the same indices $5$ and $3$.

Case 2: $n_2(G) = 3$. Since $n_2(G) = |G:N_G(S)|$ where $S \in Syl_2(G)$, the subgroup $N_G(S)$ has index $3$.

Case 3: $n_2(G) = 5$. Then as in case 2, the normalizer of any Sylow $2$-subgroup has index $5$.

Case 4: $n_2(G) = 15$. Here, I use the following theorem, proved earlier in the chapter.

Suppose that $G$ is a finite group such that $n_p(G) > 1$, and choose distinct Sylow $p$-subgroups $S$ and $T$ of $G$ such that the order of $|S \cap T|$ is as large as possible. Then $n_p(G) \equiv 1$ mod $|S:S \cap T|$.

Since $n_2(G) = 15$ is not congruent to $1$ mod $8$ or mod $4$, this theorem implies that there are $S,T \in Syl_2(G)$ such that $|S:S \cap T| = 2$, i.e. $|S \cap T| = 4$. Therefore $S\cap T \lhd S$ and $S \cap T \lhd T$, which means that $S \leq N_G(S\cap T)$ and $T \leq N_G(S\cap T)$. We write $H = N_G(S \cap T)$ for brevity. Note that $n_2(H) \equiv 1$ mod $2$, and we have $S,T \in Syl_2(H)$, so in fact $n_2(H) \geq 3$. This means that $H$ contains at least $16$ elements, so $|G:H| \leq 7$. Also, $8$ divides $|H|$, so $|G:H|$ is either $1$, $3$, or $5$. If $|G:H|$ is $3$ or $5$, then we're done: $H$ is the desired subgroup.

If $|G:H| = 1$, then $S \cap T \lhd G$. Therefore, $K = G / (S \cap T)$ is a group of order $30$. Consequently, $K$ contains subgroups of order $3$ and $5$. By the Sylow theorems, $n_3(K)$ is either $1$ or $10$, and $n_5(K)$ is either $1$ or $6$. If $n_3(K) = 10$ and $n_5(K) = 6$, then $K$ contains $20$ elements of order $3$ and $24$ elements of order $5$, but this is impossible in a group of order $30$. Therefore, either $n_3(K) = 1$ or $n_5(K) = 1$.

If $n_3(K) = 1$, then $K$ contains a normal subgroup $N$ of order $3$. Then $K/N$ is a group of order $10$, so it contains a subgroup of order $2$ (index $5$). By correspondence, $K$ also contains a subgroup of index $5$, and since $K = G / (S \cap T)$, this means that $G$ also contains a subgroup of index $5$.

Similarly, if $n_5(K) = 1$, then $K$ contains a normal subgroup $N$ of order $5$. Then $K/N$ is a group of order $6$, so it contains a subgroup of order $2$ (index $3$). By correspondence, $K$ and $G$ also contain subgroups of index $3$.

• Looks good to me! – Ravi Fernando Nov 16 '15 at 3:05

Case 4: You showed there are $S,T\in Syl_2(G)$ such that $|S\cap T|=4$ (your argument by second quoted theorem-after Case 4-is interesting, and I didn't see it earlier. But, it is nice argument.) From this, it is easy to come to your conclusion.
You have shown $S\cap T$ is normal in both $S$ and $T$, hence in $\langle S,T\rangle$.
What is order of $\langle S,T\rangle$? It contains $ST$ and $$|ST|=\frac{|S|\cdot |T|}{|S\cap T|}=16.$$ Thus, $\langle S,T\rangle$ is a subgroup of $G$ of order at least $16$, it divides $|G|$ and is divisible by $|S|=8$. So, $$|\langle S,T\rangle|=2^3.3,\,\,\, 2^3.5,\,\,\, 2^3.3.5.$$ The first two cases give a subgroup of index $5$ or $3$ respectively.
In the third case we get $S\cap T$ normal in $\langle S,T\rangle=G$. Consider quotient $G/(S\cap T)$; it is a group of order $2.3.5$.
Your above computations ensure me that you can easily prove that $G/(S\cap T)$ contains a subgroup of index $3$ and $5$, so it is true for $G$ also.
• Thanks, I didn't think to consider $\langle S,T \rangle$. However, I think everything you wrote remains valid if we use $N_G(S\cap T)$ (my $H$) in place of $\langle S,T \rangle$. I like using $|ST|$ to conclude immediately that $\langle S,T\rangle$ (or $N_G(S \cap T)$) has order at least 16. Also, that's a good point that $G/(S \cap T)$ actually contains subgroups of both indexes 3 and 5. – Bungo Nov 16 '15 at 3:31