MIT 18.701 — Lecture 4

The Correspondence Theorem, Quotient Groups, The Isomorphism Theorems, Conjugation in $S_n$

§ Proof of the Correspondence Theorem

We'll start by giving a proof of the Correspondence Theorem.

Theorem. (Correspondence Theorem) Let $f: G \to G'$ be a surjective homomorphism, and let $K = \ker(f)$ be its kernel. Consider the following collections of subgroups:

A := \{\text{subgroups of } G \text{ containing } K\} ~ \text{ and } ~ B := \{\text{subgroups of } G'\}.

Then there is a bijection from $A$ to $B$ that sends subgroups $H \in A$ to subgroups $f(H) := \{f(h): h \in H\} \in B$ .

Proof: The proof is “largely self-proving”—the statement of the theorem itself is the deep part.

Step 1. For any $H \in A$ and $H' \in B$ , we have $f(H) \in B$ and $f^{-1}(H') \in A$ .

Proof: One needs to check the following:

The set $f(H)$ contains the identity and is closed under multiplication and inverses.
The set $f^{-1}(H')$ contains the identity and is closed under multiplication and inverses.
The set $f^{-1}(H')$ contains $K$ .

The first two follow from $f$ being a homomorphism, and the third follows from the definition of $K$ . $\square$

Great! Now it remains to show $f: A \to B$ is in fact a bijection. This comes in two steps.

Step 2. For any $H \in A$ , we have $f^{-1}(f(H)) = H$ .

Proof: The fact $H \subseteq f^{-1}(f(H))$ is clear—this is true for any function $f$ and any set $H$ , in fact! So it remains to show $f^{-1}(f(H)) \subseteq H$ . The key tool for this step is the assumption that $K \subseteq H$ .

Equivalently, it remains to show that if $h \in f^{-1}(f(H))$ , then $h \in H$ .
The condition $h \in f^{-1}(f(H))$ is equivalent to the claim that $f(h) = f(h')$ for some $h' \in H$ .
What does $f(h) = f(h')$ mean? It means $h = h'k$ for some $k \in K$ , of course!
But both $h'$ and $k$ are elements of $H$ , so $h = h'k$ must also be an element of $H$ , as desired. $\square$

And here's the final step.

Step 3. For any $H' \in B$ , we have $f(f^{-1}(H')) = H'$ .

Proof: Again, $f(f^{-1}(H')) \subseteq H'$ is clear, independent of the context of $f$ and $H'$ . So it remains to show $H' \subseteq f(f^{-1}(H'))$ . The key tool for this step is the assumption that $f$ is surjective.

Equivalently, it remains to show that if $h' \in H'$ , then $h' \in f(f^{-1}(H'))$ .
Because $f$ is surjective, there exists some $h \in G$ such that $f(h) = h'$ .
Well, then for this $h$ , we have $h \in f^{-1}(H')$ . So $f(h) = h' \in f(f^{-1}(H'))$ , as desired. $\square$

Some “reading comprehension” questions:

Example. (Kernel) What goes wrong if subgroups in $A$ need not contain the kernel?

Solution: Consider a subgroup $H\subseteq G$ that does not contain $K$ . Then the subgroup $H' = f(H)$ of $G'$ contains the identity of $G'$ , so $f^{-1}(H')$ contains $K$ , but $H$ does not. So $f^{-1}(f(H)) = f^{-1}(H') \neq H$ , which is bad. $\square$

Example. (Surjectivity) What goes wrong if $f$ need not be surjective?

Solution: Then the set $B$ could be whatever it wanted to be, so $f: A \to B$ would not be surjective. $\square$

§ Quotient Groups

Recall that $H$ is a normal subgroup of $G$ (denoted $H \unlhd G$ ) if and only if $gHg^{-1} = H$ for all $g \in G$ .

Remark. Equivalently, $H \unlhd G$ if and only if $gHg^{-1} \subseteq H$ for all $g \in G$ (i.e. we only need to check one direction of inclusion). The reason is that subgroup conjugation is a bijection; $gHg^{-1}$ and $H$ always have the same size.

Also recall that every kernel $K$ of a homomorphism $f: G \to G'$ is a normal subgroup. But does the other direction hold? Must every normal subgroup $K \unlhd G$ be the kernel of some homomorphism $f: G \to G'$ ?

Theorem. (Normal Subgroups Are Kernels) For every normal subgroup $K \unlhd G$ , there exists a homomorphism $f: G \to G'$ such that $K$ is the kernel of $f$ .

Proof: Here's the key observation.

Claim. The cosets of $K$ in $G$ are closed under multiplication, where multiplication of cosets $C_1$ and $C_2$ is defined by $C_1C_2 = \{c_1c_2: c_1 \in C_1, c_2 \in C_2\}$ .

Proof: The product of $C_1 = g_1K$ and $C_2 = g_2K$ is equal to:
$(g_1K)(g_2K) = g_1(Kg_2)K = g_1(g_2K)K = (g_1g_2)K,$
which itself is a coset, as desired! $\square$

Thus, we may take $G'$ to be the group of all cosets of $K$ under the operation of coset multiplication! Then the homomorphism $f: G \to G'$ defined by $f(g) = gK$ is exactly what we need. $\blacksquare$

The group of cosets $G'$ in the proof above has a special name.

Definition (Quotient Group). Given $K \unlhd G$ , the group of cosets of $K$ is the quotient group and is denoted $G/K$ .

Remark. Suppose we try to define the product of cosets $C_1$ and $C_2$ of $H \subset G$ as follows:

“Definition”. The product of $C_1 = g_1H$ and $C_2 = g_2H$ is $C_1C_2 = (g_1g_2)H$ .

This definition is flawed! There are many different possible $g_1 \in G$ we could choose for which “ $C_1 = g_1H$ ” holds. So depending on which $g_1$ we choose as a representative of $C_1$ , we might get a different product $(g_1g_2)H$ .

The crux of the above proof is the observation that $H \unlhd G$ implies that no matter which representatives $(g_1, g_2)$ we choose, the resulting product $(g_1g_2)H$ will be consistent!

Example. Take $G = \mathbb{Z}$ and $K \unlhd G$ to be $K = n \mathbb{Z}$ . Then the quotient group $G/K = \mathbb{Z} / n \mathbb{Z}$ is isomorphic to $C_n$ .

Remark. The claim that “cosets of $K$ are closed under multiplication” is false if $K$ is not a normal subgroup. Take $G = S_3$ and $H = \{e, (1 \ 2)\}$ for a counterexample.

Example. Take the group $G = GL_n(\mathbb{R})$ and $K \unlhd G$ to be $K = SL_n(\mathbb{R})$ . Then the quotient group $G/K = GL_n(\mathbb{R})/SL_n(\mathbb{R})$ is isomorphic to $(\mathbb{R}^{\times}, \times)$ .

§ The Isomorphism Theorems

There are many isomorphism theorems, whose names are largely inconsistent across sources.

Theorem. (First Isomorphism Theorem) Let $f: G \to G'$ be a surjective homomorphism, and take $K = \ker(f)$ . Then $G/K$ is isomorphic to $G'$ under some isomorphism $\bar{f}: G/K \to G'$ .

More strongly, if we denote $\pi: G \to G/K$ , then there is a unique choice of $\bar{f}$ for which $f = \bar{f} \circ \pi$ .

Proof: The proof is similar to that of the Correspondence Theorem: it's not that deep. The constraint $f = \bar{f} \circ \pi$ implies that for any $g \in G$ , we must have:

f(g) = \bar{f} \circ \pi (g) ~ \implies ~ f(g) = \bar{f}(gK).

So there is a unique way to define $\bar{f}: G/K \to G'$ by sending $gK \in G/K$ to $f(g) \in G'$ . Some things to check:

Check #1. Is $\bar{f}$ well defined?

Proof: Yes! Suppose $g_1K$ and $g_2K$ refer to the same coset in $G/K$ ; then it suffices to show $\bar{f}(g_1K) = \bar{f}(g_2K)$ . Note that there exist $k_1, k_2 \in K$ such that $g_1k_1 = g_2k_2$ , so we may write:
$\bar{f}(g_1K) = f(g_1) = f(g_2k_2k_1^{-1}) = f(g_2) = \bar{f}(g_2K),$
as desired. (The key simplifying step $f(g_2k_2k_1^{-1}) = f(g_2)$ follows from $K = \ker(f)$ .) $\square$

Check #2. Is $\bar{f}$ an isomorphism?

Proof: Yes! The fact that $\bar{f}$ preserves multiplication can be seen as follows:
$\bar{f}(g_1K) \bar{f}(g_2K) = f(g_1)f(g_2) = f(g_1g_2) = \bar{f}(g_1g_2K) = \bar{f}((g_1K)(g_2K)).$
The fact that $\bar{f}$ is surjective follows from the surjectivity of $f$ .

And finally, the injectivity of $\bar{f}$ follows from the fact that the equation $\bar{f}(gK) = 1$ has a unique solution:
$\bar{f}(gK) = 1 ~ \iff ~ f(g) = 1 ~ \iff ~ g \in K ~ \iff ~ gK = K. ~ \square$

These two checks suffice to show that our constructed $\bar{f}$ works and is unique. $\blacksquare$

Now given any homomorphism $f: G \to G'$ , we may factor it as follows:

Note that $f$ may not be surjective, so consider the inclusion map $\mathrm{i}: \im(f) \to G'$ .
The map $f: G \to \im(f)$ is a surjective map, so we can apply the First Isomorphism Theorem on it! If $K = \ker(f)$ , then this gives us the maps $\pi: G \to G/K$ and $\bar{f}: G/K \to \im(f)$ .
Therefore, the map $f: G \to G'$ can be factored as $f = \mathrm{i} \circ \bar{f} \circ \pi$ .

See the “category theory” diagram below for a visualization of this factorization.

§ Conjugation and Normality in $S_n$

Example. Consider a permutation $\pi \in S_n$ and a cyclic permutation $\sigma = (a_1 \ a_2 \ \cdots \ a_k)$ .

What does $[ \pi \sigma \pi^{-1} ]$ look like? Well, it sends $\pi(a_i)$ to:

[ \pi \sigma \pi^{-1} ] \pi(a_i) = [ \pi \sigma ] (a_i) = \pi(a_{i + 1})

Therefore, the permutation $\pi \sigma \pi^{-1}$ can be written in cycle notation as $\pi \sigma \pi^{-1} = (\pi(a_1) \ \pi(a_2) \ \cdots \ \pi(a_k))$ . In other words, conjugating a permutation is equivalent to “relabeling” its entries.

In general, this means $H = \langle ( 1 \ 2 \ \cdots \ n ) \rangle \cong C_n \subseteq S_n$ cannot be a normal subgroup of $S_n$ for $n > 3$ . This is because $\pi H \pi^{-1}$ can be generated by any $n$ -cycle of $S_n$ , so $H$ does not remain invariant under conjugation.

§ Proof of the Correspondence Theorem

§ Quotient Groups

§ The Isomorphism Theorems

§ Conjugation and Normality in SnS_nSn​

§ Conjugation and Normality in $S_n$