51. The Ultimate Sylow Theorem Guide for Algebra Quals
(Epistemic status: Something I wish I'd had when I was a tiny undergraduate and then a yearling graduate creature, so I'm doing it for them now. I've been in the trenches, my soul's made partially of math at this point, and I helped PR - who this is partially dedicated to - do an algebra pset this week about using the Sylow theorems. Nevertheless it's been a while and I was never a dedicated algebraist; in particular, some of the "secret tech" arguments at the end have a very geometric-group-theorist flavor. If I got anything wrong I want to know. This will be useless to you unless you know a decent amount of abstract algebra already.)
A major kind of problem that shows up often for abstract algebra qualifying exams is called a "Sylow problem". They're generally of the form "Show that if \(|G| = 69\), it contains a normal subgroup" or "Classify all groups of order \(G = 42\)", and they rely on the Sylow theorems. This isn't about those theorems, though, which I won't state explicitly and will assume you already know them or can look them up. This is about my recommended workflow for straightforwardly solving a problem of the genre.
Sylow problems generally ask you to prove that some kind of group always contains a normal subgroup. A Sylow problem will always contain a number, and its numerological considerations will dictate how you approach the problem. Sometimes it won't actually be a number like 132 but rather an expression like \(p^2 q\); the same approach applies, but with more generality. In any event, you need to start by factoring the number. If it's even, your life will likely be way harder than if it's odd, but that's not always true.
Now that you've factored the number, the first thing to actually do is write down all the candidate counts for the Sylow \(p\)-groups for each prime \(p_i\). Start by remembering that every \(p\)-group has the largest order it can, which is the largest prime power which divides \(|G|\); we write \(|n_p| = p^k\). For each prime \(p_i\), Sylow's third theorem says that the count of the Sylow \(p_i\)-groups divides \(\frac{|G|}{p_i^{k_i}}\), its index in \(G\), and also that it's \(\equiv 1 \;\mathrm{mod}\; p_i\). We might already win here: the key fact to remember for this step is that since the \(p\)-groups conjugate to each other, there's a unique \(p\)-group iff that \(p\)-group is normal in \(G\).
For example, suppose we have \(|G| = 45\). We know that \(45 = 3^2 \cdot 5\), and in particular, \(\#n_3 | 5, \#n_3 \equiv 1 \;\mathrm{mod}\; 3\), and last I checked 5 is prime and also is not 1 mod 3. So there must be a single 3-group, which is normal.
Maybe this hasn't immediately worked, and we're still left with lots of candidate counts for the \(n_{p_i}\). In that case, we'll need to use more advanced techniques. My two favorite ones, which between them usually suffice, are (1) looking at how conjugation by \(G\) doesn't always scramble the \(p\)-groups differently enough and (2) counting really well.
For the former, the thing to recall is that \(G\) acts by conjugation on the \(p\)-groups, and we can get more out of that than just uniqueness iff normal subgroup. Consider any given \(g \in G\) permutes the \(p\)-groups, and this gives us a homomorphism \(\phi: G \rightarrow S_{\# n_p}\); recall that any kernel of a homomorphism is a normal subgroup of the domain group. To complete this argument - if it works, which it doesn't always - note that \(|G| > |S_{\# n_p}|\) such that \(\phi\) isn't injective.
For example, suppose we have \(|G| = 24\). We know that \(24 = 2^3 \cdot 3\), so \(\#n_2 \in \{1, 3\}\) and \(\#n_3 \in \{1, 4\}\). By assumption \(\#n_2 = 3\), because otherwise we're done. But then conjugation by elements of \(G\) gives us a map \(\phi: G \rightarrow S_3\), and since 24 > 6, we have a nontrivial kernel, which thus gives a normal subgroup anyway. Note that in particular \(|G|\) is multiple of the appropriate \(|S_n|\) - this is no mistake, given the orbit-stabilizer theorem and Lagrange's theorem, and is how you know this argument has worked.
For the latter, we can use the combination of constraints given by the size of a \(p\)-group and the number of such groups. The thing to remember here is that for \(p \neq q\), a \(p\)-group and a \(q\)-group only intersect in the identity; however, two \(p\)-groups might intersect nontrivially in a subgroup of any size strictly dividing \(p^k\). Nevertheless, counting can get results. The thing to do here is write something of the form \(1 + \sum_i \# n_i (p_i^{k_i} - 1)\), where we might replace some of the \(n_i\) by 1, if we haven't yet established that the \(p_i\)-groups intersect trivially.
For example, suppose we have \(|G| = 132\). We know that \(132 = 2^2 \cdot 3 \cdot 11\), so \(\#n_3 \in \{4, 22\}\) and \(\#n_11 = 12\) by assumption. Then we count some of \(G\)'s elements: for a contradiction, assume \(|G| = 132 \geq 1 + >(4-1) + (\geq 4) \cdot (3-1) + 12 \cdot (11-1) > 1 + 3 + 8 + 120 = 132\), a contradiction - recall that we have more than one 2-group.
For a couple of special cases, we should note that if \(|G|\) factorizes as any of \(p^2\) (and in fact any \(p^k\) for \(k > 1\)), \(pq, p^2q\) (and in fact all \(p^aq^b\)), or \(pqr\), then it's not simple, by a variety of arguments. In particular, for most of these, the same Sylow arguments work, except that now you have to quit relying on the numerology and work in full generality with arbitrary primes instead.
Let's see how this all comes together in one last example, which will also show how sometimes those arguments aren't enough and need one last technique. Suppose we have \(|G| = 90\), as shows up in the classic homework problem where you're asked to show that every group of nonprime order at most 100 which is not \(A_5\) - whose order is 60 - is non-simple. Now, \(90 = 2 \cdot 3^2 \cdot 5\), which is none of our easy patterns above, so in any case we rely on \(\# n_{p_i} | \frac{|G|}{p_i^{k_i}}\) and \(\#n_{p_i} \equiv 1 \;\mathrm{mod}\; p_i\): assuming that for all our relevant primes \(p\), \(\# n_p > 1 \), we have \(\#n_2 \in \{3, 5, 9, 15, 45\}, \#n_3 = 10, \#n_5 = 6\). The only case we can rule out using the symmetry group argument is the case where \(\#n_2 = 3\); the other possible counts of other \(p\)-groups are 5 or greater, and 120 > 90. A counting argument yields that with trivial overlap between the 3-groups, \(|G| = 1 + (\geq 3) \cdot 1 + 10 \cdot 8 + 6 \cdot 4 > 1 + 3 + 80 + 24\), which is impossible, so some pairs of the 3-groups have to intersect nontrivially. We'd be stuck, but we have one last card to play: let \(H, K\) be two 3-groups which intersect nontrivially, with \(H \cap K\) its order-3 intersection. H and K are abelian, because they have order the square of some prime, so \(H \cap K\) is normal in both, and so normal in the subset product \(HK\). What of its normalizer, \(N = N_G(H\cap K)\)? Clearly \(HK \subseteq N\), and \(N \subseteq G\) as a subgroup which also contains \(H \cap K\) as a subgroup. So \(|N| \geq 27\), and by Lagrange's theorem, \(|N|\) divides 90 and also 9 divides \(|N|\), because \(H, K\) are subgroups of \(N\). But that doesn't leave us with many options - if \(|N| = 90\), we're done, because that means that \(H\cap K\) is normal in \(G\) itself; on the other hand, if \(|N| = 45\), we have a subgroup of index 2, which is then itself normal, because then \(N\) has a single left coset and a single right coset, which are each the coset of \(N\) appropriately generated by any \(s \in G \backslash N\); we then have \(sN = Ns, sNs^{-1} = N\).
As an addendum, sometimes a more advanced version of the problem will ask you to classify all groups of a given order rather than simply proving things about whether or not a group of a given order has a normal subgroup. In this case, there's an additional step: once you've identified what normal subgroups might arise, for each normal subgroup \(N\) you look at the quotient of the group by that normal subgroup, \(G/N\). One possible group of the given order might then be a semidirect product \(N \rtimes G/N\), where \(G/N\)'s conjugation action on \(N\) gives you one such group in the classification; in particular, you look at homomorphisms from the \(G/N\) to \(Aut(N)\) and use the fact that the image of an element must have order dividing the element's order.
For example, suppose we want to classify all groups of order 21. We have \(21 = 3 \cdot 7\), with \(\#n_3 \in {1, 7\}, \#n_7 = 1\). If there's just one 3-group, we get \(\mathbb{Z}_{21} \simeq \mathbb{Z}_3 \times \mathbb{Z}_7\), because we can mod out either of the p-groups to get something cyclic. On the other hand, if there's 7 3-groups, we look at maps \(\phi: \mathbb{Z}_3 \rightarrow Aut(\mathbb{Z}_7)\), a map from an order-3 group to one of order 6. We then want to send the generator of \(\mathbb{Z}_3\) to an order-3 element of \(Aut(\mathbb{Z}_7) \simeq \mathbb{Z}_7^\times \simeq \mathbb{Z}_6^+\), that is, we get to send it to either 2 or 4, but these are the same: in both cases we get the group \( \langle a, b | a^7 = b^3 = 1, bab^{-1} = a^2 \rangle \), where the isomorphism is witnessed by the fact that 2 and 4 are multiplicative inverses in \(\mathbb{Z}_7\), so that precomposition by inversion in the \(\mathbb{Z}_3\) gives us \( \langle a, b^{-1} = b^2 | a^7 = b^{-3} = 1, bbab^{-1}b^{-1} = ba^2b^{-1} = a^4 \rangle \) - that is, the same group.
A major kind of problem that shows up often for abstract algebra qualifying exams is called a "Sylow problem". They're generally of the form "Show that if \(|G| = 69\), it contains a normal subgroup" or "Classify all groups of order \(G = 42\)", and they rely on the Sylow theorems. This isn't about those theorems, though, which I won't state explicitly and will assume you already know them or can look them up. This is about my recommended workflow for straightforwardly solving a problem of the genre.
Sylow problems generally ask you to prove that some kind of group always contains a normal subgroup. A Sylow problem will always contain a number, and its numerological considerations will dictate how you approach the problem. Sometimes it won't actually be a number like 132 but rather an expression like \(p^2 q\); the same approach applies, but with more generality. In any event, you need to start by factoring the number. If it's even, your life will likely be way harder than if it's odd, but that's not always true.
Now that you've factored the number, the first thing to actually do is write down all the candidate counts for the Sylow \(p\)-groups for each prime \(p_i\). Start by remembering that every \(p\)-group has the largest order it can, which is the largest prime power which divides \(|G|\); we write \(|n_p| = p^k\). For each prime \(p_i\), Sylow's third theorem says that the count of the Sylow \(p_i\)-groups divides \(\frac{|G|}{p_i^{k_i}}\), its index in \(G\), and also that it's \(\equiv 1 \;\mathrm{mod}\; p_i\). We might already win here: the key fact to remember for this step is that since the \(p\)-groups conjugate to each other, there's a unique \(p\)-group iff that \(p\)-group is normal in \(G\).
For example, suppose we have \(|G| = 45\). We know that \(45 = 3^2 \cdot 5\), and in particular, \(\#n_3 | 5, \#n_3 \equiv 1 \;\mathrm{mod}\; 3\), and last I checked 5 is prime and also is not 1 mod 3. So there must be a single 3-group, which is normal.
Maybe this hasn't immediately worked, and we're still left with lots of candidate counts for the \(n_{p_i}\). In that case, we'll need to use more advanced techniques. My two favorite ones, which between them usually suffice, are (1) looking at how conjugation by \(G\) doesn't always scramble the \(p\)-groups differently enough and (2) counting really well.
For the former, the thing to recall is that \(G\) acts by conjugation on the \(p\)-groups, and we can get more out of that than just uniqueness iff normal subgroup. Consider any given \(g \in G\) permutes the \(p\)-groups, and this gives us a homomorphism \(\phi: G \rightarrow S_{\# n_p}\); recall that any kernel of a homomorphism is a normal subgroup of the domain group. To complete this argument - if it works, which it doesn't always - note that \(|G| > |S_{\# n_p}|\) such that \(\phi\) isn't injective.
For example, suppose we have \(|G| = 24\). We know that \(24 = 2^3 \cdot 3\), so \(\#n_2 \in \{1, 3\}\) and \(\#n_3 \in \{1, 4\}\). By assumption \(\#n_2 = 3\), because otherwise we're done. But then conjugation by elements of \(G\) gives us a map \(\phi: G \rightarrow S_3\), and since 24 > 6, we have a nontrivial kernel, which thus gives a normal subgroup anyway. Note that in particular \(|G|\) is multiple of the appropriate \(|S_n|\) - this is no mistake, given the orbit-stabilizer theorem and Lagrange's theorem, and is how you know this argument has worked.
For the latter, we can use the combination of constraints given by the size of a \(p\)-group and the number of such groups. The thing to remember here is that for \(p \neq q\), a \(p\)-group and a \(q\)-group only intersect in the identity; however, two \(p\)-groups might intersect nontrivially in a subgroup of any size strictly dividing \(p^k\). Nevertheless, counting can get results. The thing to do here is write something of the form \(1 + \sum_i \# n_i (p_i^{k_i} - 1)\), where we might replace some of the \(n_i\) by 1, if we haven't yet established that the \(p_i\)-groups intersect trivially.
For example, suppose we have \(|G| = 132\). We know that \(132 = 2^2 \cdot 3 \cdot 11\), so \(\#n_3 \in \{4, 22\}\) and \(\#n_11 = 12\) by assumption. Then we count some of \(G\)'s elements: for a contradiction, assume \(|G| = 132 \geq 1 + >(4-1) + (\geq 4) \cdot (3-1) + 12 \cdot (11-1) > 1 + 3 + 8 + 120 = 132\), a contradiction - recall that we have more than one 2-group.
For a couple of special cases, we should note that if \(|G|\) factorizes as any of \(p^2\) (and in fact any \(p^k\) for \(k > 1\)), \(pq, p^2q\) (and in fact all \(p^aq^b\)), or \(pqr\), then it's not simple, by a variety of arguments. In particular, for most of these, the same Sylow arguments work, except that now you have to quit relying on the numerology and work in full generality with arbitrary primes instead.
Let's see how this all comes together in one last example, which will also show how sometimes those arguments aren't enough and need one last technique. Suppose we have \(|G| = 90\), as shows up in the classic homework problem where you're asked to show that every group of nonprime order at most 100 which is not \(A_5\) - whose order is 60 - is non-simple. Now, \(90 = 2 \cdot 3^2 \cdot 5\), which is none of our easy patterns above, so in any case we rely on \(\# n_{p_i} | \frac{|G|}{p_i^{k_i}}\) and \(\#n_{p_i} \equiv 1 \;\mathrm{mod}\; p_i\): assuming that for all our relevant primes \(p\), \(\# n_p > 1 \), we have \(\#n_2 \in \{3, 5, 9, 15, 45\}, \#n_3 = 10, \#n_5 = 6\). The only case we can rule out using the symmetry group argument is the case where \(\#n_2 = 3\); the other possible counts of other \(p\)-groups are 5 or greater, and 120 > 90. A counting argument yields that with trivial overlap between the 3-groups, \(|G| = 1 + (\geq 3) \cdot 1 + 10 \cdot 8 + 6 \cdot 4 > 1 + 3 + 80 + 24\), which is impossible, so some pairs of the 3-groups have to intersect nontrivially. We'd be stuck, but we have one last card to play: let \(H, K\) be two 3-groups which intersect nontrivially, with \(H \cap K\) its order-3 intersection. H and K are abelian, because they have order the square of some prime, so \(H \cap K\) is normal in both, and so normal in the subset product \(HK\). What of its normalizer, \(N = N_G(H\cap K)\)? Clearly \(HK \subseteq N\), and \(N \subseteq G\) as a subgroup which also contains \(H \cap K\) as a subgroup. So \(|N| \geq 27\), and by Lagrange's theorem, \(|N|\) divides 90 and also 9 divides \(|N|\), because \(H, K\) are subgroups of \(N\). But that doesn't leave us with many options - if \(|N| = 90\), we're done, because that means that \(H\cap K\) is normal in \(G\) itself; on the other hand, if \(|N| = 45\), we have a subgroup of index 2, which is then itself normal, because then \(N\) has a single left coset and a single right coset, which are each the coset of \(N\) appropriately generated by any \(s \in G \backslash N\); we then have \(sN = Ns, sNs^{-1} = N\).
As an addendum, sometimes a more advanced version of the problem will ask you to classify all groups of a given order rather than simply proving things about whether or not a group of a given order has a normal subgroup. In this case, there's an additional step: once you've identified what normal subgroups might arise, for each normal subgroup \(N\) you look at the quotient of the group by that normal subgroup, \(G/N\). One possible group of the given order might then be a semidirect product \(N \rtimes G/N\), where \(G/N\)'s conjugation action on \(N\) gives you one such group in the classification; in particular, you look at homomorphisms from the \(G/N\) to \(Aut(N)\) and use the fact that the image of an element must have order dividing the element's order.
For example, suppose we want to classify all groups of order 21. We have \(21 = 3 \cdot 7\), with \(\#n_3 \in {1, 7\}, \#n_7 = 1\). If there's just one 3-group, we get \(\mathbb{Z}_{21} \simeq \mathbb{Z}_3 \times \mathbb{Z}_7\), because we can mod out either of the p-groups to get something cyclic. On the other hand, if there's 7 3-groups, we look at maps \(\phi: \mathbb{Z}_3 \rightarrow Aut(\mathbb{Z}_7)\), a map from an order-3 group to one of order 6. We then want to send the generator of \(\mathbb{Z}_3\) to an order-3 element of \(Aut(\mathbb{Z}_7) \simeq \mathbb{Z}_7^\times \simeq \mathbb{Z}_6^+\), that is, we get to send it to either 2 or 4, but these are the same: in both cases we get the group \( \langle a, b | a^7 = b^3 = 1, bab^{-1} = a^2 \rangle \), where the isomorphism is witnessed by the fact that 2 and 4 are multiplicative inverses in \(\mathbb{Z}_7\), so that precomposition by inversion in the \(\mathbb{Z}_3\) gives us \( \langle a, b^{-1} = b^2 | a^7 = b^{-3} = 1, bbab^{-1}b^{-1} = ba^2b^{-1} = a^4 \rangle \) - that is, the same group.
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