Group of order 56 is not simple + Affordable Air Purifier

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Let G be a group of order 56. Show that G is not simple.

Proof:

We will use Sylow’s Theorem to show that either the 2-Sylow subgroup or 7-Sylow subgroup is normal.

|G|=2^3\cdot 7

By Sylow’s Theorem n_2\mid 7, n_2\equiv 1\pmod 2. Thus n_2=1,7.

Also, n_7\mid 8, n_7\equiv 1\pmod 7. Therefore n_7=1, 8.

If n_2=1 or n_7=1, we are done, as one of the Sylow subgroups is normal.

Suppose to the contrary n_2=7 and n_7=8.

Number of elements of order 7 = 8 x (7-1)=48

Remaining elements = 56-48=8. This is just enough for one 2-Sylow subgroup, thus n_2=1. This is a contradiction.

Therefore, a group of order 56 is simple.


Group of order 432 is not simple

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For this blog post, we shall show that a group of order 432=2^4\cdot 3^3 is not simple. We will be using several previous posts as lemmas to prove this nontrivial result.

Suppose to the contrary G is simple. By Sylow’s Third Theorem, n_3\equiv 1\pmod 3, n_3\mid 16. This means that n_3 is 1, 4 or 16.

We recall that if n_3=1, then the Sylow 3-subgroup is normal.

Let Q_1 and Q_2 be two distinct Sylow 3-subgroups of G such that |Q_1\cap Q_2| is maximum.

Using our previous lemma regarding index of intersection of Sylow subgroups, we split our analysis into three cases, the hardest of which is Case 3.

Case 1) If |Q_1\cap Q_2|=1, [Q_1:Q_1\cap Q_2]=3^3. Thus n_3=1\pmod {27}, which allows us to conclude n_3=1.

Case 2) If |Q_1\cap Q_2|=3, [Q_1:Q_1\cap Q_2]=3^2. Thus n_3=1 \pmod 9. Similarly, we can conclude n_3=1 and we are done.

Case 3) If |Q_1\cap Q_2|=9, then [Q_1:Q_1\cap Q_2]=3.

By another previous lemma regarding index of least prime divisor, Q_1\cap Q_2\trianglelefteq Q_1. Thus, Q_1\leq N_G(Q_1\cap Q_2). Thus |N_G(Q_1\cap Q_2)|=3^3\cdot 2, 3^3\cdot 2^2, 3^3\cdot 2^3,\text{or }3^3\cdot 2^4.

If N_G(Q_1\cap Q_2)=G, then Q_1\cap Q_2\trianglelefteq G which is a contradiction. Hence we suppose N_G(Q_1\cap Q_2)\neq G. Let [G:N_G(Q_1\cap Q_2)]=k. The possible values of k are k=2, 2^2, 2^3, 2^4.

Next, we use the fact that if G is a simple group and H is a subgroup of index k, then |G| divides k!.

Thus, 432\mid k!, which forces k=16.

But then Q_1=N_G(Q_1\cap Q_2) and similarly Q_2=N_G(Q_1\cap Q_2). Thus Q_1=Q_2. This is a contradiction to |Q_1\cap Q_2|=9.

Therefore all cases lead to contradiction and thus G is not simple.

Sylow subgroup intersection of a certain index + F1 Trespasser

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Back to our topic on Sylow theory…

Let G be a finite group, where q is a prime divisor of G. Suppose that whenever Q_1 and Q_2 are two distinct Sylow q-subgroups of G, Q_1\cap Q_2 is a subgroup of Q_1 of index at least q^a. Prove that the number n_q of Sylow q-subgroups of G satisfies n_q\equiv 1\pmod {q^a}.

Proof: Let \Omega=\{Q_1,Q_2\dots,Q_n\} be the set of all Sylow q-subgroups of G. Fix P=Q_k\in \Omega. Consider the group action of P acting on \Omega by conjugation.

\phi:P\times\Omega\to\Omega, \phi_x(Q_i)=xQ_i x^{-1}

By Orbit-Stabilizer Theorem, |O(Q_i)|=|P|/|N_p(Q_i)|.

We claim that N_p(Q_i)=Q_i\cap P, since any element x outside of Q_i cannot normalise Q_i, since otherwise if x \neq Q_i, xQx^{-1}=Q_i, then \langle Q_i, x\rangle will be a larger q-subgroup of G than Q_i.

Thus, |O(Q_i)|=|P|/|Q_i\cap P|\geq q^a, i.e. q^a\mid |O(Q_i)|.

|O(P)|=1.

The orbits form a partition of \Omega, thus |\Omega|=1+\sum{|O(Q_i)|}, where the sum runs over all orbits other than O(P).

Thus, n_q\equiv 1\pmod {q^a}.

 

Proof of Wilson’s Theorem using Sylow’s Theorem

Wilson’s theorem (p-1)!\equiv -1 \pmod p is a useful theorem in Number Theory, and may be proved in several different ways. One of the interesting proofs is to prove it using Sylow’s Third Theorem.

Let G=S_p, the symmetric group on p elements, where p is a prime.

|G|=p!=p(p-1)!

By Sylow’s Third Theorem, we have n_p\equiv 1\pmod p. The Sylow p-subgroups of S_p have p-1 p-cycles each.

There are a total of (p-1)! different p-cycles (cyclic permutations of p elements).

Thus, we have n_p (p-1)=(p-1)!, which implies that n_p=(p-2)!

Thus (p-2)!\equiv 1\pmod p, and multiplying by p-1 gives us (p-1)!\equiv p-1\equiv -1\pmod p which is precisely Wilson’s Theorem. 🙂

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