More on Groups: Morphisms, Normal Subgroups, etc...

Mike

2021-06-09 (Last Modified: 2021-11-06)

More on Groups: Morphisms, Normal Subgroups, etc...

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In the previous entry More of an Introduction to Group Theory, we introduced the definition of a group and also provided some lemmas to help us understand their basic structure a little better. It wasn’t as thorough as what you’d find in most Abstract Algebra books, though. But we did manage to introduce various morphisms, or maps between groups that preserve parts of their structure. Here we’ll continue with that theme.

Definition (conjugation)

Two elements of a group, $G$, $x $ and $y$ are said to be conjugate to each other if there exists $z \in G $, such that $z^{-1} \cdot x \cdot z = y $. This is equivalent to $ x \cdot z = z \cdot y $.

Similarly, two subsets $ S_1, S_2 \subseteq G $ are said to be conjugate to each other if there exists $z \in G$, such that $ z^{-1} \cdot S_1 \cdot z = S_2 $. I’ll sometimes write $ x \sim y$, or in the case of subsets, $ S_1 \sim S_2 $.

Mini-Lemma (inverse of product of inverses)

Let $x$ and $y$ be elements of the group $G$. Then $x^{-1} \cdot y^{-1} = (y \cdot x)^{-1} $.

Proof

$x^{-1} \cdot y^{-1} = (y \cdot x)^{-1} $ means, by the definition of inverse, that $x^{-1} \cdot x^{-1} \cdot (y \cdot x) = e $, the identity element of $G$. We’ve shown that we can remove, or rearrange the parentheses, so the expression on the left is equal to $ x^{-1} \cdot ( y^{-1} \cdot y ) \cdot x = e $.

QED

Mini-Lemma (inverse of inverse)

Let $x $ be an arbitrary element of a group, then $ (x^{-1})^{-1} = x $.

Proof

Let $x$ be an arbitrary element of a group, and let $ y = (x^{-1})^{-1} $. By the definition of inverse, we have $ x^{-1} \cdot y = e $, which implies that $ x \cdot x^{-1} \cdot y = x $, and that implies that $y = x $.

QED

Lemma (conjugation is transitive)

Let $ x \sim y $ and $y \sim z$, then $x \sim z $. Similarly, for subsets, $S_1 \sim S_2 $ and $S_2 \sim S_3$ implies $S_1 \sim S_3$.

Proof

Suppose $S_1 \sim S_2 $ and $S_2 \sim S_3 $, then there exists $z_1$ and $z_2 \in G$, such that $ z_1^{-1} S_1 z_1 = S_2 $, and $ z_2^{-1} S_2 z_2 = S_3 $, but then $ z_2^{-1} ( z_1^{-1} S_1 z_1 ) z_2 = S_3 $, so $ (z_2^{-1} z_1^{-1}) S_1 (z_1 z_2) = S_3 $, but since $ z_2^{-1} z_1^{-1} = (z_1z_2)^{-1} $, we have $(z_1z_2)^{-1} S_1 (z_1 z_2) = S_3$, which means that $ S_1 \sim S_3 $, as desired. This proves 2. To prove 1, just let $S_1 = \{x\}$ and $S_2 = \{y\}$.

QED

Definition (Normal Subgroup)

Let $N $, be a subgroup of $G$, if $ z^{-1} \cdot N \cdot z = N $, for every $z \in G $, then $N$ is said to be a normal subgroup of $G$. This is usually written $ N \triangleleft G $.

Now we change topics slightly, but we’ll come back around to what we’ve been talking about soon.

Definition (Kernel)

Let $A $ and $B$ be two groups. Let $ \phi : A \rightarrow B $ be a homomorphism. Then the subset of $A$ of all of the elements that $ \phi $ sends to $ e_B $, the identity element of $B$ is called the kernel of $ \phi $. This is sometimes written $ Ker(\phi) $. In set notation

$$ \begin{align} &Ker(\phi) := \\ &\{ a \in A | \phi(a) = e_B \} \\ \end{align} $$

Lemma (subgroup test)

Let $ G$ be a group and $ \emptyset \neq S \subseteq G$ be a non-empty subset of $G$. If $x \cdot y^{-1} \in S$, for any $x$ and $y \in S$, then $S \leq G$. I.e. $S$ is a subgroup of $G$.

Proof

First we verify that $e$, the identity element of $G$ must be in $S$. Let $x = y$, then $x \cdot x^{-1} \in S$, so $ e \in S $. Now we verify that each element of $S$ has an inverse in $S$. Let $y $ be an arbitrary element of $S$, then since $e \in S$, if we set $x = e$, then $ x \cdot y^{-1} = e \cdot y^{-1} = y^{-1} \in S $, so every element has an inverse in $S$. Finally we show that $S$ is closed under $ \cdot $, i.e. for any $s_1$, $s_2 \in S$, $s_1 \cdot s_2 \in S$. Let $s_1 $ and $s_2$ be two arbitrary elements of $S$. Then, since every element of $S$ has an inverse in $S$, $s_2^{-1} \in S$. But then $ s_1 \cdot (s_2^{-1})^{-1} $ must be in $S$, and since $ (s_2^{-1})^{-1} = s_2 $, we have that $s_1 \cdot s_2 \in S$, so it is closed under $ \cdot $ and is therefore a subgroup.

QED

Mini-Lemma (image of identity and inverses)

Let $A$ and $B$ be groups and $ \phi $ a homomorphism from $A$ to $B$, also let $e_A$ and $e_B$ be the identity elements of $A$ and $B$, respectively. Then

$ \phi(e_A) = e_B $.
For any $x \in A$, $ \phi(x^{-1}) = (\phi(x))^{-1} $.

Proof

Let $ \circ_A $ and $ \circ_B $ be the binary operations of $A $ and $B $. Since $ \phi$ is a homomorphism, we have for an arbitrary $ x \in A$,

$$ \begin{align} &\phi(x) &&= \\ &\phi(x \circ_A e_A ) &&= \\ &\phi(x) \circ_B \phi(e_A) \\ \end{align} $$

Thus

$$ \begin{align} &\phi(x)^{-1} \circ_B \phi(x) = \\ &\phi(x)^{-1} \circ_B \phi(x) \circ_B \phi(e_A) \\ \end{align} $$

Thus $ e_B = e_B \circ_B \phi(e_A) $ so $ e_B = \phi(e_A) $. Now for the proof of part 2. Again, let $x \in A$ be any element.

$$ \begin{align} e_B &= \phi(e_A) \\ &= \phi(x \circ_A x^{-1}) \\ &=\phi(x) \circ_B \phi(x^{-1}) \\ &\implies \\ \phi(x)^{-1} \circ_B e_B &= \phi(x)^{-1} \circ_B \phi(x) \circ_B \phi(x^{-1}) \\ &\implies \\ \phi(x)^{-1} &=\phi(x^{-1} \\ \end{align} $$

QED

Theorem (Kernels are Normal Subgroups)

Let $A$ and $B$ be two groups and let $ \phi $ be a homomorphism from $A$ to $B$, then $ Ker(\phi) \triangleleft G $.

Proof

First we prove that $Ker(\phi) $ is a subgroup. Let $x $ and $y$ be two arbitrary elements of $ Ker(\phi) $, so $ \phi(x) = \phi(y) = e_B $. Then

$$ \begin{align} &\phi(x \circ_A y^{-1}) \\ = &\phi(x) \circ_B \phi(y)^{-1} \\ = &e_B \circ_B (e_B)^{-1} \\ = &e_B \\ \end{align} $$

Therefore, $x \circ_A y^{-1} \in Ker(\phi) $. Thus by the subgroup test lemma, $ Ker(\phi) \leq G $. Now let’s prove it’s a normal subgroup. Let $z $ be an arbitrary element of $A$, and $x$ be an arbitrary element of $ Ker(\phi) $. Then

$$ \begin{align} &\phi(z^{-1} \circ_A x \circ_A z) \\ = &\phi(z^{-1}) \circ_B \phi(x) \circ_B \phi(z) \\ \end{align} $$

By the above lemmas, and since $x \in Ker(\phi) $, we have

$$ \begin{align} &\phi(z^{-1}) \circ_B \phi(x) \circ_B \phi(z) \\ = &\phi(z)^{-1} \circ_B e_B \circ \phi(z) \\ = &\phi(z)^{-1} \circ_B \phi(z) \\ = &e_B \\ \end{align} $$

Therefore

\[ z^{-1} \circ_A x \circ_A z \in Ker(\phi) \]

thus

$$ \begin{align} &z^{-1} \circ_A Ker(\phi) \circ_A z \\ &\subseteq Ker(\phi) \\ \end{align} $$

For the rest of the proof, we’ll drop the subscript $ \circ $, and instead just write $ \cdot $. So,

$$ \begin{align} &z^{-1} \cdot Ker(\phi) \cdot z \\ &\subseteq Ker(\phi) \\ \end{align} $$

implies that

$$ \begin{align} &z \cdot z^{-1} \cdot Ker(\phi) z \cdot z^{-1} \\ &\subseteq z \cdot Ker(\phi) \cdot z^{-1} \\ \end{align} $$

which means that $ Ker(\phi) \subseteq z \cdot Ker(\phi) \cdot z^{-1}$. Recall that $ z \in A$ was arbitrary, so we have

\[ z^{-1} \cdot Ker(\phi) \cdot z \subseteq Ker(\phi) \]

and

\[ Ker(\phi) \subseteq z \cdot Ker(\phi) \cdot z^{-1} \]

for all $z \in A$. So, if instead we replace all of the $z$’s with $ z^{-1} $, then the set inclusions will be reversed. Therefore the sets are all equal. This means that $ z^{-1} \cdot Ker(\phi) \cdot z = Ker(\phi) $, for any $z \in A$. Thus $ Ker(\phi) \triangleleft A $.

QED

We showed previously that cosets of a subgroup are either disjoint or equal, which means that for subgroups that have more than one element (i.e. if the subgroup isn’t just the identity element), then a coset could be represented by more than one element of the group. What this means is that there could be two elements, say, $a_1$ and $a_2$ in our group $G$, such that for the subgroup $H$, $a_1H = a_2H $. It turns out that this forms an equivalence relation on the elements of $G$. We actually dealt with an equivalence relation earlier in this entry, but we didn’t prove that it was an equivalence relation (because we hadn’t mentioned equivalence relations at all).

Definition (equivalence relation)

Let $ \equiv $ be a relation among two elements of a set, such that For all $s \in S $, $ s \equiv s$. For all $s_1$ and $ s_2 \in S$, $ s_1 \equiv s_2 \Rightarrow s_2 \equiv s_1 $. (Reflexive) For all $s_1$, $s_2$, and $s_3 \in S$, $ s_1 \equiv s_2$ and $ s_2 \equiv s_3$, then $s_1 \equiv s_3$. (Transitive)

Definition (mod a coset)

Let $a$ and $b$ be two elements of a group, $G$, and let $H$ be a subgroup of $G$. Then $a$ is said to be equal to $b$ mod $H$, written \[ a \equiv b \mod H \] if and only if the two cosets of $H$, $ aH $ and $bH$ are equal.

The syntax we’ve used for the “$ \equiv \mod H$” relation suggests that it is an equivalence relation among the elements of the group $G$, but just because we’ve used the symbol “$ \equiv \mod H $” to represent it doesn’t mean that it actually is an equivalence relation. We have to prove that by showing that $ a \equiv b \mod H$ satisfies all three of the criteria given above for an equivalence relation. Fortunately, though, it does satisfy those criteria, as we show with the following lemma.

Lemma (mod H is an equivalence relation)

$ \equiv \mod H $ is an equivalence relation.

Proof

Condition 1 and 2 are obvious. For 3, suppose $ a \equiv b \mod H $ and $ b \equiv c \mod H$ so $aH = bH$ and $bH = cH$, then by equality, $aH = cH$, which means $ a \equiv c \mod H $.

QED

Lemma (subset conjugation is an equivalence relation)

Subset conjugation $ \sim $ is an equivalence relation.

Proof

We already showed 3 (transitivity) of $ \sim $. 1 is clear by letting $z = e$, the identity. Now we prove 2. Let $ S_1 \sim S_2 $, so there is some $z \in G$, such that $ z^{-1} \cdot S_1 \cdot z = S_2 $, then, if we further conjugate by $z^{-1}$, we obtain $ z \cdot (z^{-1} S_1 \cdot z) \cdot z^{-1} = z \cdot S_2 \cdot z^{-1} $. This means that $ S_1 = z \cdot S_2 \cdot z^{-1} $, but this is just the same as $ (z^{-1})^{-1} \cdot S_2 \cdot (z^{-1}) = S_1 $, and that means that $S_2 \sim S_1$.

QED

This entry is becoming a bit long, so we’re going to continue this discussion in a future entry.