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Modern Algebra: Groups, Rings, Modules, Fields

Sam Macdonald

Section 6.2 Simple Groups

Subsection Keeping it Simple

“Everything should be made as simple as possible, but not simpler.”
―Albert Einstein

Definition 6.16. Simple Group.

A group is called simple if it has only two normal subgroups: itself and the trivial subgroup.

Example 6.17. \(\mathbb{Z}/p\) Simple.

\(\mathbb{Z}/p\) for a prime \(p\) is simple since any non-trivial element generates the whole group.

Example 6.18. No Simple Groups of Order \(12\) (Counting Elements).

Let us prove that no group or order \(12\) is simple.
Solution.
Let \(G\) be any group of order \(12\text{.}\) We will prove that \(G\) must have either a normal subgroup of order \(3\) or a normal subgroups of oder \(4\text{.}\)
Sylow’s Theorems gives that \(n_2=|\operatorname{Syl}_2(G)|\) is either \(1\) or \(3\) and \(n_3= |\operatorname{Syl}_3(G)|\) is either \(1\) or \(4\text{.}\) If either of these numbers is \(1\text{,}\) we have a unique subgroup of order \(4\) or of order \(3\text{,}\) and such a subgroup must be normal. Suppose these numbers are \(3\) and \(4\text{,}\) respectively. We deduce a contradition by “counting elements”.
In detail, say \(P_1, \dots, P_4\) are the \(4\) Sylow \(3\)-subgroups. By Lagrange’s Theorem \(P_i \cap P_j = \{e\}\) for all \(i \ne j\text{.}\) Thus the set \(T := \bigcup_{i=1}^4 P_i\) has \(9\) elements, one of which is \(e\) and the other \(8\) of which must have order \(3\text{.}\) That is, there are \(8\) elements of order \(3\) in \(G\text{.}\) But now consider the three Sylow \(2\)-subgroups \(Q_1, Q_2, Q_3\text{.}\) Each has order \(4\) and \(Q_i \cap T = \{e\}\) for all \(i\text{.}\) It follows that \(Q_i = \{e\} \cup (G \setminus T)\) for all \(i\text{,}\) and thus \(Q_1 = Q_2 = Q_3\text{,}\) a contradiction.

Remark 6.19.

In the previous example, it would not be so easy to count elements of order \(2\) and \(4\text{.}\) We do know that every element in \(S := \cup_i Q_i\) has order \(1\text{,}\) \(2\) or \(4\) (any only one has order \(1\)), but the size of this set is harder to calculate. For notice that \(Q_i \cap Q_j\) might have order \(2\text{.}\) The most one can say for sure is that \(S\) has at least \(4 + 4 - 2 = 6\) elements.

Example 6.20. No Simple Groups of Order \(80\) (Group Actions).

No group of order \(80=5 \cdot 16\) is simple.
Solution.
By way of contradiciton suppose \(G\) is simple and \(|G| = 20\text{.}\) Sylow’s Theorems gives \(|\operatorname{Syl}_2(G)| = 5\) and \(|\operatorname{Syl}_5(G)| = 16\) (since they cannot be \(1\) by the assumption that \(G\) is simple). The “counting elements trick” would work, but let’s proceed in a different way: Consider the action of \(G\) on \(\operatorname{Syl}_2(G)\) by conjugation and let
\begin{equation*} \rho: G \to S_5 \end{equation*}
be the associated homomorphism (obtained by choosing a numbering \(1, \dots, 5\) of the members of \(\operatorname{Syl}_2(G)\)). The map \(\rho\) is non-trivial since the action is transitive (part (2) of Sylow’s Theorems). But \(80\) does not divide \(120\) and so \(\rho\) cannot be injective. It follows that \(\operatorname{ker}(\rho)\) is a non-trivial, proper normal subgroup of \(G\text{,}\) a contradiction.

Example 6.21. No Simple Groups of Order \(90\) (Normalizers).

No group of order \(90\) is simple.
Solution.
By way of contradiction suppose \(G\) is a simple group of order \(90\text{.}\) Since \(90 = 2 \cdot 3^2 \cdot 5\text{,}\) we conclude from Sylow’s Theorems that
\begin{equation*} |\operatorname{Syl}_2(G)| = 1 \text{ or } 3 \text{ or } 5 \text{ or } 15 \end{equation*}
\begin{equation*} |\operatorname{Syl}_3(G)| = 1 \text{ or } 10 \end{equation*}
\begin{equation*} |\operatorname{Syl}_5(G)| = 1 \text{ or } 6. \end{equation*}
If any of these is \(1\text{,}\) then the corresonding Sylow \(p\)-subgroup is unique and hence normal, a contradiction.
Although it is not actually needed here, let’s show \(|\operatorname{Syl}_2(G)|\) cannot be \(3\) or \(5\text{.}\) For example, if it were \(5\text{,}\) then the action of \(G\) on \(\operatorname{Syl}_2(G)\) (by conjugation) gives a homomorphism \(\rho: G \to S_5\text{.}\) Since the action is transitive, this homomorphism cannot be the trivial map. Since \(90\) does not divide \(120\text{,}\) this map cannot be injective. Thus the kernel of \(\rho\) is a non-trivial, proper normal subgroup, a contradiction. In a similar way one shows \(|\operatorname{Syl}_2(G)| \ne 3\text{.}\)
We have shown that the only possibility is
\begin{equation*} |\operatorname{Syl}_2(G)| = 15, |\operatorname{Syl}_3(G)| = 10, { \text{ and } } |\operatorname{Syl}_5(G)| = 6. \end{equation*}
One is now tempted to count elements: we have \(15\) elements of order \(2\) and \(6 \cdot 4 = 24\) elements of order \(5\text{.}\) But elements of order \(3\) are not so easy to count since the various Sylow \(3\)-subgroups have nine elements each and might interect in subsets of order \(3\text{.}\) The only obvious things is that there are at least \(9 + 9 -3 = 15\) of them. This does not give enough elements to reach a contradiction.
The action of \(G\) on \(\operatorname{Syl}_g(5)\) leads to a homomorphism \(G \to S_6\text{,}\) but since \(90\) does divide \(6!\text{,}\) there is no contradiction lurking here either.
So, we must resort to something really sneaky: let \(P_1, \dots, P_{10}\) be all of the Sylow \(3\)-subgroups. Observe that if \(P_i \cap P_j = \{e\}\) for all \(i \ne j\text{,}\) then we would have \(10 \cdot 8 = 80\) elements of order \(3\) or \(9\text{.}\) This would give at least \(80+ 15 + 24\) elements, which is not possible. So, it must be the case that at least two intersect non-trivially. Without loss, let’s say \(H = P_1 \cap P_2\) is such that \(|H|= 3\text{.}\) We ponder \(N_G(H)\text{.}\) Since \(P_1\) is abelian and \(H \leq P_1\text{,}\) \(P_1 \leq N_G(H)\text{,}\) so that \(9 \mid |N_G(H)|\text{.}\) Likewise, \(P_2 \leq N_G(H)\) too, and since \(|(P_1 \cup P_2)| = 9 + 9 - 2 = 15\text{,}\) we get \(15 \leq N_G(H)\text{.}\) And of course \(|N_G(H)| \mid 90\text{.}\) We get only three possibilities:
\begin{equation*} |N_G(H)| = 18, 45 \text{ or } 90. \end{equation*}
Let us show that each is impossible:
If \(|N_G(H)| = 18\text{,}\) then we have constructed a subgroup of \(G\) of index \(5\text{.}\) The action of \(G\) on the left cosets gives a homomorphism \(G \to S_5\) that cannot be the trivial map since the action is transitive. But it cannot be injective either, and this leads to a non-trivial nomal subgroup. The exact same reasoning shows \(|N_G(H)|\) cannot be \(45\) (or we could use that it has index \(2\)). Finally, if \(|N_G(H)| = 90\text{,}\) then \(H \unlhd G\text{,}\) which is also impossible.

Proof.

Assume \(G\) is simple of order \(60\text{.}\) Let us first observe that it suffices to construct a non-trivial action of \(G\) on a set with \(5\) elements. For given such an action we obtain a non-trivial homomorphism \(g: G \to S_5\text{.}\) Since it’s non-trivial and \(G\) is simple, we must have \(\operatorname{ker}(g) = \{e\}\text{,}\) so that \(g\) is injective. Thus \(G \cong \operatorname{im}(g)\text{,}\) and \(|\operatorname{im}(g)| = 60\text{.}\) But we already know that \(A_5\) is the only subgroup of \(S_5\) of order \(60\text{,}\) and thus \(G \cong \operatorname{im}(g)=A_5\text{.}\)
So, we only need to find such an action, and one’s first guess would be the action on one of the sets of Sylow \(p\)-subgroups. By Sylow Theory and the fact that \(G\) is simple (so that no Sylow \(p\)-subgroup for \(p=2,3,5\) can be unique) we get
\begin{equation*} |\operatorname{Syl}_2(G)| \in \{3,5,15\}, |\operatorname{Syl}_3(G)| \in \{4,10\},|\operatorname{Syl}_5(G)| = 6. \end{equation*}
The next important observation is:
There is no non-trivial action of \(G\) on a set \(S\) of size at most \(4\text{.}\)
For if there was such a non-trivial action, we would get a homomorphism \(\rho: G \to S_n\) for \(2 \leq n \leq 4\) (if \(n=1\text{,}\) the only action is the trivial one). Such a homomorphism cannot be injective by order considerations. So, it’s kernel would be a proper, non-trivial normal subgroup, which do not exist.
In particular, \(|\operatorname{Syl}_2(G)| \ne 3\) and \(|\operatorname{Syl}_4(G)| \ne 4\text{,}\) so that
\begin{equation*} |\operatorname{Syl}_2(G)| \in \{5,15\},|\operatorname{Syl}_3(G)| = 10,|\operatorname{Syl}_5(G)| = 6. \end{equation*}
If \(|\operatorname{Syl}_2(G)| = 5\) then we are done, since this action is transitive and thus certainly non-trivial.
Suppose \(|\operatorname{Syl}_2(G)| = 15\) and let \(P_1, \dots, P_{15}\) be all the Sylow \(2\)-subgroups. Since \(|\operatorname{Syl}_3(G)| = 10\) and \(|\operatorname{Syl}_5(G)| = 6\text{,}\) \(G\) has \(10 \cdot 2 + 6 \cdot 4 = 44\) elements of order \(3\) or \(5\text{.}\) If \(P_i \cap P_j = \{e\}\) for all \(i \ne j\text{,}\) we would have \(15 \cdot 3\) elements of order \(2\) or \(4\text{,}\) which is far too many elements in total. At least two of these must therefore intersect non-trivially and without loss say \(H = P_1 \cap P_2\) has two elements. We consider \(K = N_G(H)\text{.}\) As in the preivious example, \(K \supseteq P_1 \cup P_2\) and hence \(|K| \geq 6\text{,}\) \(4 \mid |K|\text{,}\) and \(|K| \mid 60\text{.}\) The possibilities are
\begin{equation*} |K| = 12, 20, 60. \end{equation*}
If \(|K| = 60\) then \(H \unlhd G\text{,}\) which is impossible. If \(|K| = 20\text{,}\) then we obtain an transitive action of \(G\) on the three element set of left cosets of \(K\) in \(G\text{,}\) which as shown above is not possible. We are left with \(|K| = 12\text{.}\) Thus \(G\) acts transitively (and hence non-trivially) on the five element set of left cosets of \(K\) in \(G\text{.}\)

Remark 6.24.

The proof shows that \(G \cong A_5\) if either \(|\operatorname{Syl}_2(G)| = 5\) or \(|\operatorname{Syl}_2(G)| =15\text{,}\) but in hindsight only one of these can actually be possible. It’s actually the former: \(A_5\) has no elements of order \(4\) (since four cycles are odd permuations) and the only elements of order \(2\) are products of two disjoint transpotions, and there are \(15\) such elements. It’s not hard to see that the five \(4\)-element subgroups of \(A_5\) given by the Klein \(4\)-group \(V := \{e, (1 \, 2)(3 \, 4), (1 \, 3)(2 \, 4), (1 \, 4)(2 \, 3)\}\) and its obvious conjugates (i.e., those obtained by leaving out one of \(1,2,3,4\) instead of \(5\)) are the only Sylow \(2\)-subgoups of \(G\text{.}\)

Subsection A Wealth of Problems on Simple Groups

Exercise 6.25.

Let G be a group of order \(2^5\cdot 7^3\text{.}\) Prove that \(G\) is not simple.
Solution.
By Sylow’s Theorems we know that \(\Syl_7(G)\equiv 1\mod{7}\) and \(\Syl_7(G)|2^5=32\text{.}\) Thus our options are \(1\) and \(8\text{.}\) Suppose that \(\Syl_7(G)=8\text{.}\)
Let \(G\) act on \(\Syl_7(G)\) by conjugation, yielding the homomorphism \(\rho: G \to S_8\) via the Permutation Representation. This map is non-trivial from part (2) of Sylow’s Theorems, but \(2^5\cdot 7^3\) does not divide \(|S_8|=8!\text{,}\) and thus \(\rho\) cannot be injective. Then the kernel of this homomorphism is non-trivial, normal subgroup of \(G\) by Theorem 3.40. Thus \(G\) is not simple.

Exercise 6.26.

Let \(G\) be a group of order \(2835 = 3^4\cdot 5\cdot 7\text{.}\)
  1. Show that there are at most two options for \(n_3\text{,}\) the number of Sylow \(3\)-subgroups of \(G,\) and list them.
  2. Prove that \(G\) is not simple.
Solution.
Let \(G\) be a group of order \(2835\text{.}\)
  1. By Sylow’s Theorems we know that \(n_3=1\mod{3}\) and \(n_3|35\text{.}\) The possible options are thus \(1\) and \(7\text{.}\)
  2. Suppose by way of contradiction that \(G\) is simple. Thus \(n_3\neq 1\text{,}\) so \(n_3=7\text{.}\) Let \(G\) act on \(\Syl_3(G)\) by conjugation, yielding the homomorphism
    \begin{equation*} \rho: G\to S_7 \end{equation*}
    granted via the Permutation Representation. By (2) in Sylow’s Theorems we see that \(\rho\) is not trivial. As \(|S_7|=7!\) we see that \(|G|\not\big||S_7|\text{,}\) meaning that the \(\ker(\rho)\) is non-trivial, yielding a non-trivial normal subgroup of \(G\text{,}\) a contradiction.

Exercise 6.27.

Let \(G\) be a finite group of order \(p^2q\) with \(p < q\) prime numbers. Show that \(G\) is not a simple group.
Solution.
Let \(G\) be a finite group of order \(p^2q\) with \(p < q\) prime numbers, and suppose by way of contradiction that \(G\) is simple. By Sylow’s Theorems we have the following: - \(n_p|q\) and \(n_p\cong 1\mod{p}\text{,}\) and thus \(n_p=q\text{.}\) - \(n_q|p^2\) and \(n_q\cong 1\mod{q}\text{,}\) and thus \(n_q=p^2\text{.}\) From this information we see that there are \(p^2(q-1)\) elements of order \(q\) and \((p^2-1)q\) elements of order \(p\text{,}\) for a lovely total of \(p^2q-p^2+p^2q-q\) elements, which is too many.

Exercise 6.28.

  1. Let \(G\) be a simple group of order \(60\text{.}\) Determine the number of elements of \(G\) of order \(5\text{.}\)
  2. Show that there is no simple group of order \(30\text{.}\)
Solution.
  1. Let \(G\) be a simple group of order \(60=2^2\cdot3\cdot 5\text{.}\) By Sylow’s Theorems we know that \(n_5|12\) and that \(n_5\equiv 1\mod{5}\text{.}\) hus the options for \(n_5\) are \(1\) and \(6\text{.}\) Since \(G\) is simple we see that \(n_5=6\text{.}\) As each Sylow \(5\)-subgroup of \(G\) has \(4\) unique elements of order \(5\) and the identity we see that the number of elements or order \(5\) in \(G\) is \((5-1)\cdot 6=24\text{.}\)
  2. Suppose by way of contradiction that \(G\) is a simple group of order \(30\text{.}\) Similarly to above, \(n_5=6\text{,}\) yielding \(24\) elements of order \(5\text{.}\) Now, \(n_3|10\) and \(n_3\equiv 1\mod{3}\text{,}\) so \(n_3=10\text{,}\) yielding far too many elements to fit in \(G\text{.}\)

Exercise 6.29.

Prove that no group of order \(150\) is simple.
Solution.
Let \(G\) be a group of order \(150\) and suppose by way of contradiction that \(G\) is simple. Notice that \(150=2\cdot 3\cdot 5^2\text{.}\) By Sylow’s Theorems, we know \(|\Syl_5(G)|\equiv 1\mod{5}\) and divides \(6\text{,}\) the only options are thus \(1\) and \(6\text{.}\) Since \(G\) is simple, there must be exactly \(n_5=6\text{,}\) where \(n_5=\Syl_5(G)\text{.}\) Let \(G\) act on \(\Syl_5(G)\) by conjugation. Thus \(\rho: G\to S_6\) is a group homomorphism. Note that \(|S_6|=720\text{,}\) and that the order of \(G\) does not divide \(720\text{.}\) By Part (2) of Sylow’s Theorems this action is transitive, meaning that the kernel of \(\rho\) cannot be trivial. Thus \(\ker(\rho)\) is a nontrivial normal subgroup of \(G\text{,}\) a contradiction. Thus no group of order \(150\) is simple.
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