MIT 18.701 — Lecture 3

Kernel and Image, Cosets, Normal Subgroups, Conjugation, The Correspondence Theorem

§ Kernel and Image

Definition (Kernel and Image). Consider a homomorphism $f: G \to H$ between groups $G$ and $H$ . Then the kernel of $f$ is $\ker(f) := \{g \in G : f(g) = 1_H\}$ , and the image of $f$ is $\im (f) := \{h \in H: h = f(g) \text{ for some } g \in G\}$ .

Note that $\ker(f)$ is a subgroup of $G$ , and $\im(f)$ is a subgroup of $H$ . (Check this!)

Example. (Application of the Kernel) Take a homomorphism $f: G \to H$ , and suppose that $f(g_1) = f(g_2)$ for $g_1, g_2 \in G$ . What can we say about $g_1$ and $g_2$ ?

Solution: Observe the following:

f(g_1) = f(g_2) ~ \iff ~ f(g_1g_2^{-1}) = 1 ~ \iff ~ g_1g_2^{-1} \in \ker(f) ~ \iff ~ g_1 = kg_2 ~ \text{ for some } k \in \ker(f).

If we multiply by the left inverse instead of the right inverse in the first step, we also find:

f(g_1) = f(g_2) ~ \iff ~ f(g_2^{-1}g_1) = 1 ~ \iff ~ g_2^{-1}g_1 \in \ker(f) ~ \iff ~ g_1 = g_2k \text{ for some } k \in \ker(f).

Consolidating these results together, we have:

\boxed{\begin{cases}f(g_1) = f(g_2) ~ \iff ~ g_1 = kg_2 \text{ for some } k \in \ker(f) \\ f(g_1) = f(g_2) ~ \iff ~ g_1 = g_2k \text{ for some } k \in \ker(f)\end{cases}}

Example. Take the homomorphism $f: (\mathbb{C}, +) \to (\mathbb{C}^{\times}, \cdot)$ defined by $f(z) := e^z$ . Then the image of $f$ is $\im(f) = \mathbb{C}^{\times}$ (i.e. $f$ is surjective), and the kernel of $f$ is $\ker(f) = 2\pi i \cdot \mathbb{Z}$ .

Example. Take the homomorphism $f: GL_n(\mathbb{R}) \to \mathbb{R}^{\times}$ defined by $f(M) = \det M$ . Then the image of $f$ is $\im(f) = \mathbb{R}^{\times}$ (i.e. $f$ is surjective), and the kernel of $f$ is $\ker(f) = SL_n(\mathbb{R})$ .

Example. Take the homomorphism $f: S_n \to \{-1, 1\}$ sending permutations $\pi \in S_n$ to the sign of $\pi$ . The kernel of $f$ must be a subgroup of $S_n$ ; this subgroup is the alternating group $A_n$ , consisting of all even permutations in $S_n$ .

Example. Let $G$ be a group, and fix an element $g \in G$ . Then we may take the homomorphism $f: (\mathbb{Z}, +) \to G$ defined by $f(n) := g^n$ . In this case,

The image of $f$ is the subgroup generated by $\{g\}$ . We denote this by $\im(f) = \langle g \rangle$ .
The kernel of $f$ is $\ker(f) = k\mathbb{Z}$ , where $g$ has order $k$ . (If $g$ has infinite order, then $\ker(f) = \{0\}$ .)

§ Cosets

Example. Recall that $3\mathbb{Z}$ is a subgroup of $\mathbb{Z}$ . This subgroup can be “copied-and-translated” twice to form the subsets $3\mathbb{Z} + 1$ and $3\mathbb{Z} + 2$ in $\mathbb{Z}$ . (Note that these subsets are not subgroups!)

Let's give a name to these “copied-and-translated” subsets associated with a subgroup: cosets.

Definition (Cosets). Let $H$ be a subgroup of $G$ . Then a left coset of $H$ is a subset $gH := \{gh : h \in H\} \subseteq G$ for some $g \in G$ . Similarly, a right coset of $H$ is a subset $Hg := \{hg : h \in H\} \subseteq G$ for some $g \in G$ .

The phrase “coset” (without further specification) should be assumed to mean “left coset”.

Theorem. (Coset Properties) Let $H$ be a subgroup of $G$ . The cosets of $H$ in $G$ satisfy the following properties:

Any two cosets of $H$ have the same size.
Any two distinct cosets are disjoint.

Proof: To prove the first fact, simply note that the size of any coset $gH$ is equal to $|H|$ . This is because the natural map $T: H \to gH$ defined by $T(h) = gh$ is a bijection. $\blacksquare$

To prove the second fact, consider two distinct cosets $C_1 = g_1H$ and $C_2 = g_2H$ . Suppose, for the sake of contradiction, that these two cosets are not disjoint. Then there exists $h_1, h_2 \in H$ such that $g_1h_1 \in C_1$ and $g_2h_2 \in C_2$ are the same; that is, $g_1h_1 = g_2h_2$ . Equivalently, $g_1 = g_2h_2h_1^{-1}$ . However, this implies

C_1 = g_1H = (g_2h_2h_1^{-1})H = g_2(h_2h_1^{-1}H) = g_2H = C_2.

So in fact, we have shown that if there is just a single pair of elements $g_1h_1$ and $g_2h_2$ in $C_1$ and $C_2$ that match, then we can show the entirety of $C_1$ and $C_2$ must match completely! $\blacksquare$

Here's an equivalent, more intuitive rephrasing of the above theorem:

Theorem. (Cosets form Partitions) Let $H$ be a subgroup of $G$ . Then the set of all left cosets of $H$ partitions the elements of $G$ into disjoint subsets of size $|H|$ .

Note that all of the above theorems hold equally true if every instance of “left” is replaced with “right”. However, the partitioning of $G$ formed by left cosets usually does not equal the partitioning formed by right cosets!

Example. Consider the group $G = (\mathbb{R}^2, +)$ and a subgroup $H := \{(a, 2a): a \in \mathbb{R}\}$ . Then cosets of $H$ look like translated copies of lines parallel to $y = 2x$ . These lines tile the plane $\mathbb{R}^2$ .

Definition (Index). The number of cosets of $H$ in $G$ is the index of $H$ in $G$ , denoted $[G:H]$ . Note that the theorem above implies $|G| = [G:H] \cdot |H|$ .

Theorem. (Lagrange's Theorem) Let $H$ be a (finite) subgroup of a (finite) group $G$ . Then $|H|$ divides $|G|$ .

Proof: Follows from the equation $|G| = [G:H] \cdot |H|$ , along with the fact that $[G:H]$ is an integer. $\blacksquare$

Lagrange's Theorem is very powerful! Here's an interesting corollary of the above theorem.

Theorem. (Groups of Prime Order) Any group of prime order must be cyclic.

Proof: Suppose $G$ is a group of prime order $p$ . Pick some $g \in G$ that is not the identity. Then $H = \langle g \rangle$ is a subgroup of $G$ . Importantly, $|H| > 1$ (since $g$ is not the identity), and $|H|$ divides $p$ . So it must be that $|H| = p$ , meaning that $G$ and $H$ are the same group. And since $H$ was a cyclic group, $G$ must be cyclic, too.

§ Normal Subgroups

How do cosets and homomorphisms relate?

Example. (Application of the Coset) Take a homomorphism $f: G \to H$ , and suppose that $f(g_1) = f(g_2)$ for $g_1, g_2 \in G$ . What can we say about $g_1$ and $g_2$ ?

Solution: Recall the following result we derived earlier:

\boxed{\begin{cases}f(g_1) = f(g_2) ~ \iff ~ g_1 = kg_2 \text{ for some } k \in \ker(f) \\ f(g_1) = f(g_2) ~ \iff ~ g_1 = g_2k \text{ for some } k \in \ker(f)\end{cases}}

We can now use cosets to rephrase the above statement as follows:

\boxed{\begin{cases}f(g_1) = f(g_2) ~ \iff ~ g_1 \text{ and } g_2 \text{ are in the same right coset of } \ker(f). \\ f(g_1) = f(g_2) ~ \iff ~ g_1 \text{ and } g_2 \text{ are in the same left coset of } \ker(f). \end{cases}} ~~~~~ (\bigstar)

The above motivates the following definition:

Definition (Normal). A subgroup $N \subseteq G$ is normal if $gN = Ng$ for all $g \in G$ .

This is because the above result $(\bigstar)$ immediately implies the following theorem:

Theorem. (Normal Kernel) For a homomorphism $f: G \to H$ , its kernel $K = \ker(f)$ is a normal subgroup of $G$ .

Example. Consider $G = S_3$ and a subgroup $N = \{e, (1 \ 2)\}$ . Then:

The left cosets of $N$ are $N = \{e, (1 \ 2)\}$ , $(1 \ 3)N = \{(1 \ 3), (1 \ 2 \ 3)\}$ , and $(2 \ 3)N = \{ (2 \ 3), (1 \ 3 \ 2)\}$ .
The right cosets of $N$ are $N = \{e, (1 \ 2)\}$ , $N(1 \ 3) = \{(1 \ 3), (1 \ 3 \ 2)\}$ , and $N(2 \ 3) = \{(2 \ 3), (1 \ 2 \ 3)\}$ .

Thus, $N$ is not normal, since (e.g.) $N(1 \ 3) \neq (1 \ 3)N$ . In other words, the partitioning of $G$ formed by the left cosets of $N$ is different from the partitioning of $G$ formed by the right cosets of $N$ .

Remark. By corollary, there is no homomorphism $f: G \to H$ to any other group $H$ such that $\ker(f) = N$ .

Example. If $G$ is abelian, then every subgroup $N \subseteq G$ is normal. This is because:

gN = \{gn : n \in N\} = \{ng : n \in N\} = Ng.

Example. If $N$ is a subgroup of $G$ such that $[G:N] = 2$ , then $N$ must be normal. This is because any partitioning of $G$ into cosets—whether they be left cosets or right cosets—must look like $G = N \ \sqcup \ (G \setminus N)$ . So the (two) left cosets of $N$ must match the (two) right cosets of $N$ .

§ Conjugation

Definition (Conjugation). Given elements $a, b \in G$ , the conjugation of $a$ by $b$ is defined to be $bab^{-1}$ . More generally, if $H$ is a subgroup of $G$ , then the conjugation of $H$ by $a$ is $aHa^{-1}$ .

It turns out (easy to check) that if $H$ is a subgroup, then $aHa^{-1}$ is also always a subgroup. Thus, we can think of conjugation by $g \in G$ as an action that sends subgroups of $G$ to subgroups of $G$ .

Theorem. (Normality by Conjugation Condition) A subgroup $N \subseteq G$ is normal if and only if $N$ does not change under conjugation by any element $g \in G$ .

Proof: Recall that $N$ is normal if and only if $gN = Ng$ for all $g \in G$ . Equivalently, $gNg^{-1} = N$ for all $g \in G$ , which may directly be read as: the conjugation of $N$ by $g$ equals $N$ for any $g \in G$ . $\blacksquare$

Theorem. (Intersecting Normal Subgroups) If $N_1$ and $N_2$ are normal subgroups of $G$ , then their intersection $N_1 \cap N_2$ is also normal.

Proof: We argue using the formulation of normality in terms of conjugation described above. For any $g \in G$ ,

\begin{align*}g(N_1 \cap N_2)g^{-1} = \ & \{gng^{-1} : n \in N_1 \cap N_2\} \\ = \ & \{gng^{-1} : n \in N_1\} \cap \{gng^{-1}: n \in N_2\} \\ = \ & (gN_1g^{-1}) \cap (gN_2g^{-1}) \\ = \ & N_1 \cap N_2 \text{ since } N_1 \text{ and } N_2 \text{ are normal.}\end{align*}

Thus, $N_1 \cap N_2$ remains unchanged by conjugation by $g$ , so $N_1 \cap N_2$ is normal. $\blacksquare$

§ Correspondence Theorem

Here's a useful theorem that we won't prove today.

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 $f(H) := \{f(h): h \in H\} \in B$ .

Example. (Correspondence Theorem Application #1) Consider the determinant homomorphism $f = \det: G \to G'$ , with $G = GL_n(\mathbb{R})$ and $G' = \mathbb{R}^{\times}$ . Then $K = \ker(f) = SL_n(\mathbb{R})$ . What does the Correspondence Theorem say?

Solution: The set $B := \{\text{subgroups of } \mathbb{R}^{\times}\}$ is easier to analyze; for example, $\mathbb{Q}^{\times}$ is a subgroup in $B$ .

By the Correspondence Theorem, the subgroup $\mathbb{Q}^{\times}$ must correspond to a new, quirky subgroup of $GL_n(\mathbb{R})$ containing $K = SL_n(\mathbb{R})$ that we didn't know of before. This subgroup is:

f^{-1}(\mathbb{Q}^{\times}) = \{M \in GL_n(\mathbb{R}): \det M \in \mathbb{Q}^{\times}\} \in A.

There are many, many other subgroups of $GL_n(\mathbb{R})$ that live between $K$ and $G$ .

Example. (Correspondence Theorem Application #2) Take $G = \mathbb{Z}$ and $G' = C_n = \langle g \rangle$ , and define the homomorphism $f: G \to G'$ by $f(k) = g^k$ . What does the Correspondence Theorem say?

Solution: Observe that $\ker(f) = n \mathbb{Z}$ . Then:

A := \{\text{subgroups of } G \text{ containing } K\} = \{d\mathbb{Z} : d \mid n\}.

It then follows that we can classify the set of all subgroups of $G' = C_n$ :

B := \{ \text{subgroups of } G'\} = f(A) = \{f(d\mathbb{Z}) : d \mid n\} = \{ \langle g^d \rangle : d \mid n \}.

So in fact, this tells us every subgroup of $C_n$ is of the form $\langle g^d \rangle$ for some $d \mid n$ .