31 Quotient Groups and Homomorphisms

#todo/school : Do problems 3.1 numbered 1, 3, 4, 5, 7, 14ab, 22, 24, 27, 31, 36, 41 📅 2025-02-09 ✅ 2025-02-09
#todo/school : Read 3.1 (pp. 89-99). You left off noting Left (and Right) Cosets form a Partition in G for pp.94 📅 2025-02-09 ✅ 2025-02-07

1

Question

Let $φ : G \to H$ be a Homomorphism (or Group Morphism) and let $E$ be a Subgroup of $H$ . Prove that $φ^{- 1} (E) \leq G$ , namely that the preimage or pullback of a subgroup under a homomorphism is a subgroup.
If $E ⊴ H$ prove that $φ^{- 1} (E) ⊴ G$ . Deduce that $Ker (φ) ⊴ G$ . (see Normal Subgroup)

Proof

$(φ^{- 1} (E) \leq G)$ : Let $a, b^{- 1} \in φ^{- 1} (E)$ . We just have to show that $a b^{- 1} \in φ^{- 1} (E)$ . By the definition of Fiber (Group Theory), namely $φ^{- 1} (E) = {g \in G | φ (g) \in E}$ then there are elements $e_{1}, e_{2} \in E$ such that $φ (a) = e_{1}$ and $φ (b^{- 1}) = e_{2} = φ (b)^{- 1}$ . Notice that:

\begin{aligned} φ (a b^{- 1}) & = φ (a) φ (b)^{- 1} \\ = e_{1} e_{2} \\ \in G \end{aligned}

Thus choose $e = e_{1} e_{2} \in G$ . Thus we found some element $e \in G$ such that $φ (a b^{- 1}) = e \in E$ , so then $a b^{- 1} \in φ^{- 1} (E)$ as desired.

$(E ⊴ H \Rightarrow φ^{- 1} (E) ⊴ G)$ : Suppose $E ⊴ H$ . Already we have $φ^{- 1} (E) \leq G$ so to show that $φ^{- 1} (E) ⊴ G$ we want to show one of the Equivalences for Normal Subgroups, namely that for any $g \in G$ and $a \in φ^{- 1} (E)$ that $g a g^{- 1} \in φ^{- 1} (E)$ .

Let $g \in G$ and $a \in φ^{- 1} (E)$ , so then there is some $e \in E$ such that $φ (a) = e$ . Construct $h = φ (g) \in H$ . Notice that since $E ⊴ H$ then for all $h \in H, e \in E$ then $h e h^{- 1} \in E$ . But we have this since our constructed $h \in H$ and thus:

h e h^{- 1} \in E

Thus there is some $e^{'} \in E$ where $h e h^{- 1} = e^{'}$ . But look:

\begin{aligned} h e h^{- 1} \in E & ⟺ φ (g) e φ (g^{- 1}) \in E \\ ⟺ φ (g) φ (a) φ (g^{- 1}) \in E \\ ⟺ φ (g a g^{- 1}) \in E \\ ⟺ g a g^{- 1} \in φ^{- 1} (E) \end{aligned}

as we desired.

$(Ker (φ) ⊴ G)$ : We already have $Ker (φ) \leq G$ since it is a clear subset and any $g_{1}, g_{2}^{- 1} \in Ker (φ)$ has it that $φ (g_{1}) = φ (g_{2}^{- 1}) = 1_{H}$ so then $1_{H} = φ (g_{1}) φ (g_{2}^{- 1}) = φ (g_{1} g_{2}^{- 1})$ so then $g_{1} g_{2}^{- 1} \in Ker (φ)$ (showing it's a subgroup).

Consider that $Ker (φ) = φ^{- 1} (1_{H})$ . Thus replacing all the $E$ 's with ${1_{H}}$ we trivially know ${1_{H}} ⊴ H$ so then by our findings then $Ker (φ) ⊴ G$ .

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3

Question

Let $A$ be an Abelian Group and let $B$ be a Subgroup of $A$ . Prove that $\frac{A}{B}$ (the Quotient Group) is also abelian. Give an example of a non-abelian group $G$ containing proper Normal Subgroup $N$ such that $\frac{G}{N}$ is abelian.

Proof

(Abelian part): Here $\frac{A}{B}$ is the set of left-cosets of $B$ in $A$ , or just that for any $a \in A$ then $a B = {a b | b \in B}$ are all elements of $\frac{A}{B}$ .

Let $X, Y \in \frac{A}{B}$ be arbitrary. Then $X = x B$ for $x \in A$ and $Y = y B$ where $y \in A$ . We want to show that $\frac{A}{B}$ is abelian, which is true if $X Y = Y X$ . This is true using the Coset Product since:

\begin{aligned} X Y & = (x B) (y B) \\ = (x y) B \\ = (y x) B & (A is abelian \Rightarrow x y = y x) \\ = (y B) (x B) \\ = Y X \end{aligned}

Thus $\frac{A}{B}$ is abelian.

(Example): Let $G = D_{8} = ⟨ r, s : r^{4} = 1, s^{2} = 1, r s = s r^{3} ⟩$ which is non-abelian. Consider the normal subgroup $N = {1, r^{2}}$ which is normal because it is abelian. Notice then that $\frac{G}{N}$ is abelian as it is ${g n : n \in N} = {g r^{2 i} : 0 \leq i < 2}$ .

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4

Question

Prove that in the Quotient Group $\frac{G}{N}$ , $(g N)^{α} = g^{α} N$ for all $α \in Z$ .

Proof

We are assuming, via the definition of quotient group, that $N \leq G$ .

For the base case of our induction notice using Coset Product:

(g N)^{2} = (g N) (g N) = (g g) N = g^{2} N

continuing on gives all $α \in Z^{+}$ cases.

Notice for $α = 0$ then $(g N)^{0} = N = g^{0} N$ as $N$ is the identity for the quotient group.

Doing a similar induction on $α$ over $Z^{-}$ will work as long as we show the base case that $(g N)^{- 1} = g^{- 1} N$ . Notice that:

(g N)^{- 1} (g N) = N = (g^{- 1} g) N

by the definition of being the inverse. But what is the inverse of $(g N) = g N$ ? Namely notice that:

1 N = (g^{- 1} g) N = (g^{- 1} N) (g N)

Thus then:

(g N)^{- 1} (g N) = (g^{- 1} N) (g N)

and using Cancellation Theorem (right cancellation):

(g N)^{- 1} = g^{- 1} N

as we desired.

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5

Question

Use the preceding exercise to prove that the Order (Groups) of the element $g N$ in $\frac{G}{N}$ (see Quotient Group) is $n$ , where $n$ is the smallest positive integer such that $g^{n} \in N$ (and $g N$ has infinite order if no such positive integer exists).
Give an example to show that the order of $g N$ in $\frac{G}{N}$ may be strictly smaller than the order of $g$ in $G$ .

Proof

(Proof): Say $n$ is the smallest positive integer such that $g^{n} \in N$ . Then using our previous finding then:

(g N)^{n} = g^{n} N = N

as $g^{n} \in N$ . Thus the order of $g N$ by definition being in $\frac{G}{N}$ must be $n$ .

If such an $n$ didn't initially exist, then $\forall n \in N (g^{n} \notin N)$ . Assume for contradiction that some $n \in N$ exists such that $(g N)^{n} = N$ (implying that there's a finite order for $g N$ ). Then:

(g N)^{n} = g^{n} N

but since $g^{n} \notin N$ then $g^{n} N \neq N$ , which is a contradiction (contradicts the order of $g N$ being $n$ ). Thus such an $n$ cannot exist so $| g N |$ is infinite.

(Example): Consider $G = S_{4}$ . Let $x$ be a generator for $G$ , and take $N = ⟨ x^{2} ⟩ = {1, x^{2}}$ . Now $| x^{2} | = 2$ in $G$ (since $(x^{2})^{2} = x^{4} = 1$ ) but $x^{2} N = 1 N$ has order 1 in $\frac{G}{N}$ (since $x^{2} N = 1 N$ ).

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7

Question

Define $π : R^{2} \to R$ where $(x, y) \mapsto x + y$ . Prove that $π$ is a surjective Homomorphism (or Group Morphism) and describe the Kernel and fibers of $π$ geometrically.

Proof

( $π$ is surjective): Let $r \in R$ be arbitrary. Construct the element $(r, 0) \in R^{2}$
such that $π ((r, 0)) = r + 0 = r$ . Thus $π$ is arbitrary.

(Homomorphism (or Group Morphism)): Let $\vec{x}, \vec{y} \in R^{2}$ be arbitrary. Say $\vec{x} = (x_{1}, x_{2})$ and $\vec{y} = (y_{1}, y_{2})$ . Notice that (using the product of vectors to give a vector with corresponding entries added together, and using addition over $R$ ):

\begin{aligned} π (\vec{x} \vec{y}) & = π (\vec{x} + \vec{y}) \\ = π (x_{1} + x_{2}, y_{1} + y_{2}) \\ = x_{1} + x_{2} + y_{1} + y_{2} \\ = (x_{1} + y_{1}) + (x_{2} + y_{2}) \\ = π (\vec{x}) + π (\vec{y}) \\ = π (\vec{x}) π (\vec{y}) \end{aligned}

The kernel $Ker (π) = {\vec{x} \in R^{2} | π (\vec{x}) = 0}$ would be the set of vectors:

\vec{z} = (z, - z) \forall z \in R

Since $π (\vec{z}) = z + (- z) = 0$ so clearly $\vec{z} \in Ker (π)$ . If any other vector didn't follow this then it wouldn't be in the kernel. These are the vectors that point in the same direction of $(1, - 1)$ , creating a line that intercepts the origin.

Further for any specific $r \in R$ then the fiber above $r$ would be:

π^{- 1} (r) = {\vec{z} \in R^{2} | π (\vec{z}) = r}

Notice if we consider any $\vec{z} = (z_{1}, z_{2})$ here then if we want $π (\vec{z}) = r$ then:

\Rightarrow z_{1} + z_{2} = r \Rightarrow \vec{z} = (z_{1}, r - z_{1})

Dropping $z = z_{1}$ then any $z \in R$ gives a vector $\vec{z} = (z, r - z) = z (1, - 1) + (0, r)$ . Thus, geometrically, any fiber above $r$ would be the lines pointing in the direction $(1,)$ with y-intercept at $r$ .

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14ab

Question

Consider the additive Quotient Group $\frac{Q}{Z}$ :
a. Show that every coset of $Z$ in $Q$ contains exactly one representative $q \in Q$ in the range $0 \leq q < 1$ .
b. Show that every element of $\frac{Q}{Z}$ has finite order but that there are elements of arbitrarily large order.

Proof

a. We'll just consider left cosets as the right coset case is largely the same. Take any coset of $Z$ in $Q$ , denoted $q Z$ . We want to show that if $q_{1} Z = q_{2} Z$ then $q_{1} = q_{2} \in [0, 1)$ .

Suppose $q_{1} Z = q_{2} Z$ . Since the Left (and Right) Cosets form a Partition in G and since $q_{1} Z = q_{2} Z$ then $q_{2}^{- 1} q_{1} \in Z$ . Then there is some integer $m \in Z$ where:

m = q_{2}^{- 1} + q_{1}

Note

We are dealing with the additive quotient group, so what looks like multiplication of the elements of $Q$ here is actually their addition.

Going a bit back, realize that if $q_{1}, q_{2} \notin [0, 1)$ then they equal some integer plus a fraction that is in that range, and then we'd get rational numbers that are in that range:

q_{1} = n_{1} + \underset{\in [0, 1)}{\underset{⏟}{q_{1}^{'}}}, q_{2} = n_{2} + q_{2}^{'} \Rightarrow m = q_{1}^{'} - q_{2}^{'} = (n_{1} - n_{2}) + \underset{\in [0, 1)}{\underset{⏟}{(q_{1}^{'} - q_{2}^{'})}} \Rightarrow m^{'} = q_{1}^{'} - q_{2}^{'}

Thus we know that at the very least we can represent our $q$ 's with $q_{1}, q_{2} \in [0, 1)$ . For contradiction assume that $q_{1} > q_{2}$ . Notice that $m = 0$ because $m = q_{1} - q_{2}$ which are in the set $[0, 1)$ with $q_{1} > q_{2}$ . That means that $q_{1} = q_{2}$ which is a contradiction, so then $q_{1} ≯ q_{2}$ . A similar argument shows $q_{1} ≮ q_{2}$ so then $q_{1} = q_{2}$ , as desired.

b. Let $q Z \in \frac{Q}{Z}$ . Using (a) then $q \in [0, 1)$ and is unique. Then have $q = \frac{a}{b}$ where $b \neq 0$ and $a < b$ . Notice that here since we are dealing with an additive group then our Order (Groups) comes from repeated multiplication, meaning we want to see when $n q \in Z$ for some integer $n$ .

Notice though that $b q = a \in Z$ exactly! Thus, then $q$ has order $b$ , which must be some positive natural number.

Since $q Z$ was arbitrary, each coset represented by $q$ has a finite order given by the denominator of $q$ . We can always make this $b$ arbitrarily large.

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22

Question

a. Prove that if $H$ and $K$ are Normal Subgroups of a Group $G$ then their intersection $H \cap K$ is also a normal subgroup of $G$ .
b. Prove that the intersection of an arbitrary nonempty collection of normal subgroups of a group is a normal subgroup (you cannot assume the collection is countable).

Proof

See 21 Subgroups#10 for showing how $H \cap K$ (and an arbitrary intersection of them) is also a subgroup.

a. We already know $H \cap K \leq G$ so we need to show that $g (H \cap K) = (H \cap K) g$ for all $g \in G$ (since then we have Equivalences for Normal Subgroups (1)). Let $g \in G$ . Notice that for some $p \in g (H \cap K)$ then:

\begin{aligned} p \in g (H \cap K) & ⟺ \exists h \in H \cap K (p = g h) \\ ⟺ p = h g & (H ⊴ G \land K ⊴ G) \\ ⟺ \exists h \in H \cap K (p = h g) \\ ⟺ p \in (H \cap K) g \end{aligned}

where that middle step comes from the fact that if $h \in H$ then you use $H ⊴ G$ and if not then $h \in K$ so you use $K ⊴ G$ .

Overall, since $p$ was arbitrary then $g (H \cap K) = (H \cap K) g$ showing normalcy, and thus $H \cap K ⊴ G$ .

b. Say such a collection $A = {H_{λ} : λ \in Λ}$ is something we are intersecting over. We want to see if:

A = ⋂_{λ \in Λ} H_{λ}

is a normal subgroup of $G$ . As we showed before it's a subgroup so all we have to show is that for any $g \in G$ that we have:

g (⋂_{λ \in Λ} H_{λ}) = (⋂_{λ \in Λ} H_{λ}) g

Now then we have some $p \in g (⋂_{λ \in Λ} H_{λ})$ iff:

\begin{aligned} p \in g (⋂_{λ \in Λ} H_{λ}) & ⟺ \exists h \in ⋂_{λ \in Λ} H_{λ} (p = g h) \\ ⟺ p = h g \\ ⟺ \exists h \in ⋂_{λ \in Λ} H_{λ} (p = h g) \\ ⟺ p \in (⋂_{λ \in Λ} H_{λ}) g \end{aligned}

where the abelianness, like in (a), comes from the fact that $h \in ⋂_{λ \in Λ} H_{λ}$ means there any $H_{λ} \in A$ has $H_{λ} ⊴ G$ , so then since $h$ is in at least one of these $H_{λ}$ 's then $H_{λ} ⊴ G$ and thus $p = h g = g h$ .

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24

Question

Prove that if $N ⊴ G$ (see Normal Subgroup) and $H$ is any subgroup of $G$ then $N \cap H ⊴ H$ .

Proof

Notice that $N \cap H \subseteq H$ . Notice also if we let $g_{1}, g_{2} \in N \cap H$ then we can show that $g_{1}^{- 1} g_{2} \in N \cap H$ since $g_{1}, g_{2} \in H$ so then by $N, H$ being a subgroup of $G$ then $g_{1}^{- 1}, g_{2} \in N \cap H$ and thus their product is in $N \cap H$ . Thus $N \cap H \leq H$ .

Now we want to do the Equivalences for Normal Subgroups, so we'll show that $h (N \cap H) = (N \cap H) h$ for all $h \in H$ . Let $h \in H$ . Then for any $p$ :

\begin{aligned} p \in h (N \cap H) & ⟺ \exists n \in N \cap H (p = h n) \\ ⟺ p = n h & (n \in N ⊴ G ⟺ n g = g n \forall g \in G) \\ ⟺ \exists n \in N \cap H (p = n h) \\ ⟺ p \in (N \cap H) h \end{aligned}

where to clarify for the middle step, because $n \in N ⊴ G$ then that's iff $n g = g n$ for all $g \in G$ . But $H \leq G$ implies that $h \in H \subseteq G$ thus $h \in G$ ! Thus this applies and we can make the swap.

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27

Question

Let $N$ be a finite subgroup of a group $G$ . Show that $g N g^{- 1} \subseteq N$ iff $g N g^{- 1} = N$ (this is essentially showing one of the equivalences for Equivalences for Normal Subgroups). Deduce that $N_{G} (N) = {g \in G | g N g^{- 1} \subseteq N}$ (see Normalizer).

Proof

Clearly the reverse direction is true so well only check the forward direction. Suppose for any $g \in G$ that $g N g^{- 1} \subseteq N$ for some finite $N \leq G$ . We want to show that $N \subseteq g N g^{- 1}$ for that same $g \in G$ .

Notice that since $N$ is a finite subgroup then $| g N g^{- 1} | \leq | N |$ by $g N g^{- 1} \subseteq N$ . Assume for contradiction that instead $N \supset g N g^{- 1}$ . Then by that same reasoning then $| g N g^{- 1} | < | N |$ . However, we can create a bijection $φ : N \to g N g^{- 1}$ where $n \mapsto g n g^{- 1}$ . To show this is a bijection:

(injective): Let $x, y \in N$ and suppose $φ (x) = φ (y)$ . Then $g x g^{- 1} = g y g^{- 1}$ , where with some Cancellation Theorem then $x = y$ .
(surjective): Let $g n g^{- 1} \in g N g^{- 1}$ . Choose element $n$ such that $φ (n) = g n g^{- 1}$ as desired.
This implies that $| N | = | g N g^{- 1} |$ , contradicting $| g N g^{- 1} | < | N |$ . Thus then we must have $N \subseteq g N g^{- 1}$ , so then overall we have $g N g^{- 1} = N$ as desired.

Notice using the definition of Normalizer with our equivalence gives the needed deduction.

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31

Question

Prove that if $H \leq G$ and $N$ is a Normal Subgroup of $H$ then $H \leq N_{G} (N)$ . Deduce that $N_{G} (N)$ is the largest subgroup of $G$ in which $N$ is normal (ie: is the joining of all subgroups $H$ for which $N ⊴ H$ ).

Proof

( $H \leq N_{G} (N)$ ): Suppose $H \leq G$ and $N ⊴ H$ .

First to show $H \subseteq N_{G} (N)$ . Let $h \in H$ . We want to show that $h \in N_{G} (N) = {g \in G | g N g^{- 1} = N}$ , which is true iff $h N h^{- 1} = N$ . But since $N ⊴ H$ then by the Equivalences for Normal Subgroups then we automatically get that.

Second, clearly $H$ will be closed under binary operations and inverses by being a subgroup of $G$ , so we don't need to check that.

Notice that $N$ is a normal subgroup of itself, so then $N ⊴ N_{G} (N)$ (replace $H$ with $N$ from our finding above, then add that $N$ itself is already normal). Since $N$ was arbitrary, then the largest normal $N$ we can choose must be when we use $N_{G} (N)$ as our normal subgroup.

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36

Question

Prove that if $\frac{G}{Z (G)}$ is a Cyclic Group then $G$ is an Abelian Group.

In proof terms, you'll want to show that if $\frac{G}{Z (G)}$ is cyclic with generator $x Z (G)$ then show that every element of $G$ can be written in the form $x^{a} z$ for some integer $a \in Z$ and some element $z \in Z (G)$ . For a refresher on the notation see Centralizer.

Proof

Let $\frac{G}{Z (G)}$ be a quotient group that's also a cyclic group and centralizer $Z (G) = C_{G} (G) = {g \in G | g h = h g \forall h \in G}$ . This quotient group is the set of all left-cosets:

\frac{G}{Z (G)} = {g Z (G) : g \in G}

Now, let's do the proof. Because $\frac{G}{Z (G)}$ is cyclic then it can be generated by one single element $x \in G$ such that:

\frac{G}{Z (G)} = {x^{i} Z (G) : i \in Z}

To start showing that $G$ is abelian, let $g_{1}, g_{2} \in G$ be arbitrary. Notice that $g_{1} Z (G), g_{2} Z (G) \in \frac{G}{Z (G)}$ . Thus then $\exists i, j \in Z$ such that:

g_{1} Z (G) = x^{i} Z (G), g_{2} Z (G) = x^{j} Z (G)

Since $g_{1} Z (G) = x^{i} Z (G)$ then by Left (and Right) Cosets form a Partition in G then we must have $x^{- i} g_{1} \in Z (G)$ (and a similar thing for $x^{- j} g_{2} \in Z (G)$ . If $x^{- i} g_{1} = z_{1}$ and $x^{- j} g_{2} = z_{2}$ then by each $z_{1}, z_{2} \in Z (G)$ then the $z$ 's will commute with any of the $g$ 's and $x$ 's (since either is in $G$ ). Thus:

x^{- i} g_{1} = z_{1} \Rightarrow g_{1} = x^{i} z_{1} : g_{2} = x^{j} z_{2}

Furthermore the $z$ 's will commute with each other, as they each are also in $G$ . Thus:

\begin{aligned} g_{1} g_{2} & = (x^{i} z_{1}) (x^{j} z_{2}) \\ = x^{i} x^{j} z_{1} z_{2} & (x^{n} z_{1} = z_{1} x^{n} \forall n \in Z) \\ = x^{j} x^{i} z_{1} z_{2} \\ = x^{j} x^{i} z_{2} z_{1} & (z_{1} z_{2} = z_{2} z_{1}) \\ = (x^{j} z_{2}) (x^{i} z_{1}) & (x^{n} z_{2} = z_{2} x^{n} \forall n \in Z) \\ = g_{2} g_{1} \end{aligned}

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41

Question

Let $G$ be a Group. Prove that $N = ⟨ x^{- 1} y^{- 1} x y | x, y \in G ⟩$ is a Normal Subgroup of $G$ and $\frac{G}{N}$ is an Abelian Group (here $N$ is called the commutator subgroup of $G$ ).

Proof

( $N$ is normal): Clearly already $N \leq G$ . For the normal part, let $g \in G$ and $x^{- 1} y^{- 1} x y \in N$ (for some $n \in N$ ). Then we have:

\begin{aligned} g (x^{- 1} y^{- 1} x y) g^{- 1} & = g x^{- 1} (g^{- 1} g) y^{- 1} (g^{- 1} g) x (g^{- 1} g) y g^{- 1} \\ = (g x^{- 1} g^{- 1}) (g y^{- 1} g^{- 1}) (g x g^{- 1}) (g y g^{- 1}) \\ \in N \end{aligned}

where by Equivalences for Normal Subgroups then $N$ must be normal. Notice that raising our generator to various powers won't change it being in $N$ as $N$ is closed under these powers.

( $\frac{G}{N}$ is abelian): Let $z \in N$ be arbitrary. Then:

x^{- 1} y^{- 1} x y = z \Rightarrow x y = y x z

Now notice that:

\begin{aligned} a \in x y N & ⟺ a = x y z_{0} \exists z_{0} \in N \\ ⟺ a = y x z z_{0} \\ ⟺ a \in y x N \end{aligned}

suggesting that $x y N = y x N$ for any $x, y \in G$ . Thus for any arbitrary $a N, b N \in \frac{G}{N}$ then:

(a b) N = (b a) N \Rightarrow a b = b a

since $N$ is normal.

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