## 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.

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 $|HK|=\frac{|H|}{|H\cap K|}\cdot |K|$.

## 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

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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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