Advanced Method for Proving Normal Subgroup

For beginners in Group Theory, the basic method to prove that a subgroup H is normal in a group G is to show that “left coset = right coset”, i.e. gH=Hg for all g\in G. Variations of this method include showing that ghg^{-1}\in H, gHg^{-1}=H, and so on.

This basic method is good for proving basic questions, for example a subgroup of index two is always normal. However, for more advanced questions, the basic method unfortunately seldom works.

A more sophisticated advanced approach to showing that a group is normal, is to show that it is a kernel of a homomorphism, and thus normal. Thus one often has to construct a certain homomorphism and show that the kernel is the desired subgroup.

Example: Let H be a subgroup of a finite group G and [G:H]=p, where p is the smallest prime divisor of |G|. Show that H is normal in G.

The result above is sometimes called “Strong Cayley Theorem”.

Proof: Let G act on G/H by left translation.

G\times G/H\to G/H, (g,xH)\to gxH.

This is a group action since 1\cdot xH=xH, and g_1(g_2\cdot xH)=g_1g_2xH=(g_1g_2)\cdot xH.

This action induces a homomorphism \sigma:G\to S_{G/H}\cong S_p. Let g\in\ker\sigma. \sigma(g)(xH)=xH for all xH\in G/H, i.e. gxH=xH for all x\in G. In particular when x=1, gH=H. This means that g\in H. So we have \ker\sigma\subseteq H.

Suppose to the contrary \ker\sigma\neq H, i.e. [H:\ker\sigma]>1. Let q be a prime divisor of [H:\ker\sigma].

We also have

[G:\ker\sigma]=[G:H][H:\ker\sigma]=p[H:\ker\sigma]

By the First Isomorphism Theorem, G/ker\sigma\cong\text{Im}\ \sigma\leq S_p. By Lagrange’s Theorem, [G:\ker\sigma]\mid p!, i.e. p[H:\ker\sigma]\mid p!. This implies [H:\ker\sigma]\mid(p-1)!. Finally, q\mid(p-1)! implies q\leq p-1<p.

However, q\mid [H:\ker\sigma] implies q\mid[G:\ker\sigma]=\frac{|G|}{|\ker\sigma|} which implies q\mid|G|.

This is a contradiction that p is the smallest prime divisor of |G|. Thus, H=\ker\sigma and therefore H is a normal subgroup.

This proof is pretty amazing, and hard to think of without any hints.

|HK|=|H||K|/|H intersect K|

Finally, the LaTeX path not specified problem has been solved by WordPress!

This post is about how to prove that |HK|=\frac{|H||K|}{|H\cap K|}, where H and K are finite subgroups of a group G.

A tempting thing to do is to use the “Second Isomorphism Theorem”, HK/H\cong K/(H\cap K). However that would be a serious mistake since the conditions for the Second Isomorphism Theorem are not met. In fact HK may not even be a group.

The correct way is to note that HK=\bigcup_{h\in H}hK.

Therefore |HK|=|K|\times |\{hK:h\in H\}|. For h_1,h_2\in H, we have:

\begin{aligned}h_1K=h_2K&\iff h_1h_2^{-1}\in K\\    &\iff h_1h_2^{-1}\in H\cap K\\    &\iff h_1(H\cap K)=h_2(H\cap K)    \end{aligned}

Therefore |\{hK:h\in H\}|=|\{h(H\cap K):h\in H\}|, i.e. the number of distinct cosets h(H\cap K). Since H\cap K is a subgroup of H, applying Lagrange’s Theorem gives the number of distinct cosets h(H\cap K) to be \frac{|H|}{|H\cap K|}.

Thus, we have |HK|=\frac{|H|}{|H\cap K|}\cdot |K|.


Undergraduate Math Books

 

Order of a^k (Group Theory)

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Today we will revise some basic Group Theory. Let G be a group and a\in G. Assume that a has finite order n. Find the order of a^k where k is an integer.

Answer: \displaystyle|a^k|=\frac{n}{(n,k)}, where (n,k)=\gcd(n,k).

Proof:

Our strategy is to prove that m=\frac{n}{(n,k)} is the least smallest integer such that (a^k)^m=1.

Now, we have \displaystyle a^{k\cdot\frac{n}{(n,k)}}=(a^n)^{\frac{k}{(n,k)}}=1. Note that k/(n,k) is an integer and thus a valid power.

Suppose to the contrary there exists c<\frac{n}{(n,k)} such that a^{kc}=1.

Since a has finite order n, we have n\mid kc, which leads to \displaystyle\frac{n}{(n,k)}\mid\frac{k}{(n,k)}\cdot c. Note that \frac{n}{(n,k)} and \frac{k}{(n,k)} are relatively prime.

Thus \frac{n}{(n,k)}\mid c, which implies that \frac{n}{(n,k)}\leq c which is a contradiction. This proves our result. 🙂

A solvable group that has a composition series is necessarily finite

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Let G be a solvable group. We prove that if G has a composition series, then G has to be finite. (Note that this is sort of a converse to “A finite group has a composition series.”)

Let G=G_0\geq G_1\geq \dots\geq G_n=1 be a composition series of G, where each factor G_i/G_{i+1} is simple.

Since G_i and G_{i+1} are solvable (every subgroup of a solvable group is solvable), the quotient G_i/G_{i+1} is also solvable.

We can prove that G_i/G_{i+1} is abelian. Since (G_i/G_{i+1})'\trianglelefteq G_i/G_{i+1}, by the fact that the factor is simple, we have (G_i/G_{i+1})'=1 or G_i/G_{i+1}.

If (G_i/G_{i+1})'=G_i/G_{i+1}, then this contradicts the fact that G_i/G_{i+1} is solvable. Thus (G_i/G_{i+1})'=1 and G_i/G_{i+1} is abelian.

Key step: G_i/G_{i+1} is simple and abelian, G_i/G_{i+1}\cong\mathbb{Z}_{p_i} for some prime p_i.

Since |G_{n-1}|=p_{n-1}, so we have that |G_{n-2}|=|G_{n-2}/G_{n-1}||G_{n-1}|=p_{n-2}p_{n-1}. By induction, |G_i|=p_i p_{i+1}\dots p_{n-1}.

|G|=|G_0|=p_0p_1\dots p_{n-1}. Thus G is finite.

 

Aut(Z_n): Automorphism Group of Z_n

Do check out our list of Recommended Undergraduate Math Books!

We prove that Aut(\mathbb{Z}_n)\cong (\mathbb{Z}/n\mathbb{Z})^*, also known as U(n) (easier to type).

Define \Psi: Aut(\mathbb{Z}_n)\to U(n) by \Psi(\phi)=\phi (1).

First we show that it is a homomorphism:

\begin{aligned}\Psi(\phi_1 \circ \phi_2)&=\phi_1(\phi_2(1))\\    &=\phi_1 (1+1+\cdots +1)\ \ \ (\phi_2 (1) \text{ times})\\    &=\phi_1 (1)+\phi_1 (1)+\cdots +\phi_1 (1)\ \ \ (\phi_2 (1) \text{ times})\\    &=\phi_2 (1) \cdot \phi_1 (1)\\    &=\Psi (\phi_2)\cdot \Psi (\phi_1)\\    &=\Psi (\phi_1) \cdot \Psi (\phi_2)\ \ \ \text{since} (\mathbb{Z}/n\mathbb{Z})^* \text{ is abelian.}    \end{aligned}

Next we show that it is injective:

\Psi (\phi) =1

Thus, \phi (1)=1.

Let x\in \mathbb{Z}_n.

\phi (x)=x\phi (1)=x\cdot 1=x.

Thus, the only automorphism that maps to 1 is the identity.

Thus, \ker \Psi is trivial.

Finally, we show that it is surjective.

Let x\in (\mathbb{Z}/n\mathbb{Z})^*. Consider \phi such that \phi (0)=0, \phi (1)=x, \phi (i)=ix, …, \phi (n-1)=(n-1)x.

We claim that \phi is an automorphism of \mathbb{Z}_n.

Firstly, we need to show that \{0,1,2,\cdots, n-1\}=x\{0, 1, 2, \cdots, n-1\}. This is because \gcd (x,n)=1. Hence if q is the order of x, i.e. qx\equiv 0 \pmod n, then n\vert qx, which implies that n\vert q which implies that q is at least n. Since the order of x is also at most n, q=n.

Finally, we have \phi(a+b)=(a+b)x=ax+bx=\phi (a)+\phi (b) and thus we may take \phi as the preimage of x.

Hence \Psi is surjective.

This is a detailed explanation of the proof, it can be made more concise to fit in a few paragraphs!

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