33 The Isomorphism Theorems

#todo/school : Section 3.3: 3, 4, 7, 8 📅 2025-02-16 ✅ 2025-02-17

3

Question

Prove that if $H$ is a Normal Subgroup of $G$ (ie: $H ⊴ G$ ) of prime Index (Groups) $p$ then for all $K \leq G$ either:
i. $K \leq H$
ii. $G = H K$ and $| K : K \cap H | = p$

Proof

If (i) then we are done so suppose $K ≰ H$ .

First using Lagrange's Divisibility Theorem of Order of Subgroups:

| K : K \cap H | = \frac{| K |}{| K \cap H |}

and doing it again using the above information:

| G : H | = \frac{| G |}{| H |} = p

Then because Prime Order Groups are Cyclic and Z_p then $\frac{G}{H} ≅ Z_{p}$ . This implies that there is some isomorphism $φ$ where:

g H \mapsto n mod p

Now because $K ≰ H$ then $\exists k \in K$ such that $k \notin H$ . Now since $k \in G$ then since $k \notin H$ then $φ (k H)$ is some value that's not the identity. Then because $p$ is prime, then $⟨ φ (k H) ⟩$ generates all of $Z_{p}$ which has order $p$ . Thus $| φ (k H) | = p$ and by being a bijection then $| k | = p$ .

Thus $⟨ k ⟩$ has order $p$ . Notice that it must generate all of $K$ since $| G | = p | H |$ (if it didn't, then $| G |$ would be a larger order). Thus then $⟨ k ⟩ = K$ . Consequently, then $G = H K$ .

Lastly, notice that $K \cap H = {1}$ as if $x \in K \cap H$ then $x = k^{n}$ for some $0 \leq n \leq p$ and also $x \in H$ , so then $n = 0$ , so then $x = k^{0} = 1$ . Thus any $x = 1$ so then $K \cap H = {1}$ . Thus $| K \cap H | = 1$ , so:

| K : K \cap H | = \frac{| K |}{1} = | K | = | ⟨ k ⟩ | = p

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4

Question

Let $C$ be a Normal Subgroup of the group $A$ and let $D$ be a normal subgroup of $B$ . Prove that $(C \times D) ⊴ (A \times B)$ and $\frac{A \times B}{C \times D} ≅ (\frac{A}{C}) \times (\frac{B}{D})$ .

Proof

We have $C ⊴ A$ and $D ⊴ B$ . It's clear that $(C \times D)$ and $(A \times B)$ are Groups, and furthermore it's easy to see how $(C \times D) ⊴ (A \times B)$ since:

$(C \times D) \subseteq (A \times B)$ and is a group, so $(C \times D) \leq (A \times B)$ by construction.
For the normal part, let $(a, b) \in (A \times B)$ . Then since $C ⊴ A$ and $D ⊴ B$ then:

\begin{aligned} (a, b) (C \times D) & = {(a, b) (c, d) : (c, d) \in (C \times D)} \\ = {(a c, b d) : (c, d) \in (C \times D)} \\ = {(c a, d b) : (c, d) \in (C \times D)} & (C ⊴ A, D ⊴ B) \\ = {(c, d) (a, b) : (c, d) \in (C \times D)} \\ = (C \times D) (a, b) \end{aligned}

For the isomorphism part, consider the map $φ : \frac{A \times B}{C \times D} \to (\frac{A}{C} \times \frac{B}{D})$ where $(a, b) (C \times D) \mapsto (a C, b D)$ . Notice:

\begin{aligned} Ker (φ) & = {(a, b) \in \frac{A \times B}{C \times D} : (a C, b D) = (C, D)} \\ = {(a, b) \in A \times B : a C = C, b D = D} \\ = {(a, b) : a \in C, b \in D} \\ = C \times D \end{aligned}

Thus using The First Isomorphism Theorem of Groups, The Fundamental Theorem for Group Morphisms:

\frac{A \times B}{C \times D} ≅ φ (A \times B) = (\frac{A}{C} \times \frac{B}{D})

where the last step is gotten since $φ$ is surjective as if we let $(a C, b D)$ in the codomain then we can always construct $(a, b)$ such that $φ (a, b)$ maps to it.

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7

Question

Let $M, N ⊴ G$ such that $G = M N$ . Prove that $\frac{G}{(M \cap N)} ≅ (\frac{G}{M} \times \frac{G}{N})$ .

Proof

This implies we should try The Second Isomorphism Theorem (The Diamond Iso. Theorem). Namely because $M, N ⊴ G$ then automatically we have $N_{G} (M) = N_{G} (N) = G$ , so then $M \leq N_{G} (N)$ and $N \leq N_{G} (M)$ , allowing us to use the theorem. Thus we get:

G = M N ≅ \frac{M N N}{N} ≅ \frac{M N}{M \cap N}

But we can show that $G ≅ (\frac{G}{M} \times \frac{G}{N})$ using the map $φ : G \to (\frac{G}{M} \times \frac{G}{N})$ such that $g \mapsto (g M, g N)$ . Then notice:

\begin{aligned} Ker (φ) & = {g \in G : φ (g) = 1} \\ = {g \in G : (g M, g N) = (M, N)} \\ = {g \in G : g M = M; g N = N} \\ = {g \in G : g \in M; g \in N} \\ = {g \in G : g \in M \cap N} \\ = M \cap N \end{aligned}

Thus using the The First Isomorphism Theorem of Groups, The Fundamental Theorem for Group Morphisms:

\frac{G}{Ker (φ)} = \frac{G}{M \cap N} ≅ φ (G) = (\frac{G}{M} \times \frac{G}{N})

because $φ$ is surjective.

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8

Question

Let $p$ be a prime and let $G$ be the Group of $p$ -power roots of $1$ in $C$ . Prove that the map $z \mapsto z^{p}$ is a surjective Homomorphism (or Group Morphism). Deduce that $G$ is isomorphic to a proper quotient of itself.

Proof

Let $φ$ be this map:

Surjectivity: Let $y \in G$ . Then $y = e^{2 π k i / p^{n}}$ for some $n \in N, k \in Z$ . Construct $z = e^{2 π \frac{k i}{p^{n + 1}}}$ so then $φ (z) = {(e^{2 π \frac{k i}{p^{n + 1}}})}^{p} = e^{2 π \frac{k i}{p^{n}}} = y$ as needed.
Homomorphism: Let $z, w \in C$ . Then $φ (z w) = (z w)^{p} = z w z w \dots z w z w = z^{p} w^{p} = φ (z) φ (w)$ .

Now to show that $G ≅ \frac{G}{Ker (φ)}$ , use The First Isomorphism Theorem of Groups, The Fundamental Theorem for Group Morphisms. Note that $Ker (φ)$ may not be trivial (ex: if $p = 2$ then $Ker (φ) = {1, - 1} \neq {1}$ ), but we can always make this argument.

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