## 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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