LEC ALG2 5

The Correspondence Theorem

Let $\varphi :G\to \mathcal{G}$ be a homomorphism, and let $H<G$. Consider the restriction $\varphi {|}_H:H\to \mathcal{G}$. Note that

• $\ker (\varphi {|}_H)=(\ker \varphi )\cap H$
• We know that the image of $\varphi {|}_H$ divides both $|H|$ and $|\mathcal{G}|$. If $|H|$ and $|\mathcal{G}|$ have no common factors, then $\varphi (H)=\{1_{\mathcal{G}}\}$, and $H$ is contained in the kernel.

Theorem 1(Proposition).

Let $\varphi :G\to \mathcal{G}$ be a homomorphism with kernel $K$ and let $\mathcal{H}$ be a subgroup of $\mathcal{G}$. Denote the inverse image ${\varphi }^{-1}(\mathcal{H})$ by $H$. Then

1. $H<G$, $H\supset K$.

2. If $\mathcal{H}◃\mathcal{G}$, then $H◃G$.

3. If $H◃G$ and $\varphi$ is surjective, $\mathcal{H}◃\mathcal{G}$

Proof.

2. If$\mathcal{H}◃\mathcal{G}$, let $x\in H$ and $g\in G$. Then, $\varphi (gx{g}^{-1})=\varphi (g)\varphi (x)\varphi (g^{-1})\in \mathcal{H}$, i.e, $gx{g}^{-1}\in H$.
3. Let $a\in \mathcal{H}$, $b\in \mathcal{G}$. Since $\varphi$ is surjective, there exist $x\in H$, $g\in G$ such that $\varphi (x)=a$ and $\varphi (g)=b$. Since $H$ is normal, $gx{g}^{-1}\in H$, so $\varphi (gx{g}^{-1})=ba{b}^{-1}\in \mathcal{H}$.□

Theorem 2(The correspondence theorem).

Let $\varphi :G\to \mathcal{G}$ be a surjective group homomorphism with kernel $K$. There is a bijective correspondence between subgroups of $\mathcal{G}$ and subgroups of $G$ that contain $K$:

A subgroup $H<G$ that contains $K$ $\leftrightarrow$ its image $\varphi (H)$ in $\mathcal{G}$
A subgroup $\mathcal{H}<\mathcal{G}$ $\leftrightarrow$ its inverse image ${\varphi }^{-1}(\mathcal{H})$ in $G$.

If $H$ and $\mathcal{H}$ are corresponding subgroups, then $H◃G⟺\mathcal{H}◃\mathcal{G}$. Also, $|H|=|\mathcal{H}||K|$.

Proof.

Note that if$H<G$ containing $K$ (and for any subgroup in general), $\varphi (H)<\mathcal{G}$, and for any $\mathcal{H}<\mathcal{G}$, ${\varphi }^{-1}(\mathcal{H})<G$ and contains $K$, from the previous proposition. We now have to show the bijectivity of the correspondence. $\varphi ({\varphi }^{-1}(\mathcal{H}))=\mathcal{H}$ is true for any surjective map, so is $H\subset {\varphi }^{-1}(\varphi (H))$. It remains to be shown that $H\supset {\varphi }^{-1}(\varphi (H))$. Let $x$ be an element of ${\varphi }^{-1}(\varphi (H))$. By definition of the inverse image, $\varphi (x)\in \varphi (H)$, say $\varphi (x)=\varphi (a)$. Then, $a^{-1}x$ is in the kernel $K$, and since $H$ contains $K$, $a^{-1}x$ is in $H$. Since both $a$ and $a^{-1}x$ are in $H$, $x$ is in $H$ too.

The remaining assertion follows from the previous proposition.□

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Composition series and solvable groups

Definition 3(Definition).

A (finite or infinite) group $G$ is called simple if $|G|>1$ and the only normal subgroups of $G$ are $1$ and $G$.

Definition 4(Definition).

In a group $G$, a sequence of subgroups

$$1=N_0◃N_1◃\cdots ◃N_{k-1}◃N_k=G$$

is called a composition series if $N_{i+1}/N_i$ is a simple group, $0\le i\le k-1$. If the above sequence is a composition series, the quotient groups $N_{i+1}/N_i$ are called composition factors of $G$.

Theorem 5(Theorem).

If $G$ is a finite group with order greater than 1, then there exists a composition series of $G$, and the composition factors of $G$ are unique up to permutation and isomorphism.

Theorem 6(Theorem).

If $G$ is a simple group of odd order, then $G\cong Z_p$ for some prime $p$.

Definition 7(Definition).

A group $G$ is solvable if there is a chain of subgroups

$$1◃G_0◃G_1◃\cdots ◃G_s=G$$

such that $G_{i+1}/G_i$ is abelian for $0\le i\le s-1$.

The following generalization of Sylow’s theorem characterizes finite solvable groups:

Theorem 8(Theorem).

The finite group $G$ is solvable iff for every divisor $n$ of $|G|$ such that $\left(n,\frac{|G|}{n}\right)=1$, $G$ has a subgroup of order $n$.

Theorem 9(Theorem).

For $N◃G$, if $N$ and $G/N$ are solvable, then $G/N$ is solvable.