16 Homomorphisms, Isomorphisms

#todo/school : Do problems 1, 2, 5, 11, 13, 14, 16, 17, 18, 20 📅 2025-01-17 ✅ 2025-01-17

1

Question

Let $φ : G \to H$ be a Homomorphism (or Group Morphism).
a. Prove that $φ (x^{n}) = φ (x)^{n}$ for all $n \in Z^{+}$
b. Do (a) for $n = - 1$ and deduce that $φ (x^{n}) = φ (x)^{n}$ for all $n \in Z$ .

Proof

a. We'll do this by induction. Clearly $n = 1$ works here. Let the inductive hypothesis be that $φ (x^{n - 1}) = φ (x)^{n - 1}$ and any lower powers. Now notice that since $φ$ is a homomorphism then:

φ (x^{n}) = φ (x^{n - 1} \cdot x) = φ (x)^{n - 1} φ (x) = φ (x)^{n}

thus completing the inductive step.

b. We'd do an induction on $n \in Z^{+}$ and proving:

φ (x^{- n}) = φ (x)^{- n}

To do this, first consider the base case $n = - 1$ :

φ (x^{- 1}) = ?

Notice we want to prove that the left is $φ (x)$ 's inverse, which we can verify:

φ (x^{- 1}) φ (x) = φ (x^{- 1} \cdot x) = φ (1) = 1

since $φ$ is a Homomorphism (the first step, the other order works the same way) and the last identity step holds true since for any $x \in G$ (using Cancellation Theorem):

φ (x \cdot 1) = φ (x) φ (1) = φ (x) \Rightarrow φ (1) = 1

Continuing on, you'd make the inductive step just like we did for (a).

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2

Question

If $φ : G \to H$ is an Isomorphism, prove that $| φ (x) | = | x |$ for all $x \in G$ . Deduce that any two isomorphic groups have the same number of elements of order $n$ for each $n \in Z^{+}$ . Is the result true if $φ$ is only assumed to be a homomorphism?

Proof

Let $x \in G$ . If we let $| x | = n$ then:

x^{n} = 1

Now notice using 16 Homomorphisms, Isomorphisms#1 (since being an isomorphism implies it's a homomorphism) that:

φ (x)^{n} = φ (x^{n}) = φ (1) = 1

Thus $| φ (x) | = n = | x |$ as desired.

Showing that $G, H$ have the same number of elements of order $n$ for each $n \in Z^{+}$ , let $n \in Z^{+}$ . Say $G$ only has $k_{n}$ elements of order $n$ . Then $φ$ is a bijection that shows that there are at least $k_{n}$ elements of the same order $n$ (use the first part). But notice that if there were any more elements, then we can consider those extra elements $s_{1}, \dots$ and apply $φ^{- 1}$ to them and get an element $φ^{- 1} (s_{1}) \in G$ we missed (ie: $| s_{1} | = n$ which is a contradiction), so then we must have exactly $k_{n}$ elements for that order $n$ .

Notice that this doesn't work for homomorphisms (where that $φ^{- 1}$ existence may not be guarunteed). For example, the map $h : Z \to \frac{Z}{3 Z}$
where $h (u) = u mod 3$ is a homomorphism, but not an isomorphism. Furthermore, $Z$ has no elements of order say $2$ but $\frac{Z}{3 Z}$ has one (namely $\overset{―}{1}$ ).

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5

Question

Prove that the additive groups $R$ and $Q$ are not isomorphic.

Proof

Clearly we need to use the Irrationality of sqrt(2), so we want to try to make an isomorphism such that $f (\sqrt{2}) = \sqrt{2}$ to break it. Namely, assume for contradiction that such an isomorphism $f : R \to Q$ exists.
Then:

f (x + y) = f (x) + f (y)

and $f^{- 1}$ exists. Consider $f (x)$ where $x$ is an integer. Then:

f (x) = f (\sum_{i = 1}^{x} 1) = \sum_{i = 1}^{x} f (1) = x f (1) = x

So then $f$ maps each integer to itself. Similarly for rational numbers $\frac{a}{b} \in Q$ :

b \cdot f (\frac{a}{b}) = \sum_{i = 1}^{b} f (\frac{a}{b}) = f (\sum_{i = 1}^{b} \frac{a}{b}) = f (a) = a

So:

f (\frac{a}{b}) = \frac{a}{b}

So if we consider $f (x)$ when $x \in I$ then clearly if $f$ mapped $x$ to itself then that would contradict $x \notin Q$ . But if it didn't map to itself then it would have to map to a rational number that's already been mapped (which contradicts $f$ being a bijection).

Thus we have a contradiction, so $f$ cannot exist, so $R$ and $Q$ are not isomorphic.

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11

Question

Let $A, B$ be Groups. Prove that $A \times B ≅ B \times A$ .

Proof

For clarity say $A$ has the operation $\cdot$ and $B$ has the operation $⋆$ .

Define the map $φ : A \times B \to B \times A$ such that for all $a \in A, b \in B$ :

φ (a, b) = (b, a)

We'll first show that $φ$ is a valid Homomorphism (or Group Morphism). Namely, let $c \in A, d \in B$ . Then:

\begin{aligned} φ ((a, b) (c, d)) & = φ (a \cdot b, c ⋆ d) \\ = (c ⋆ d, a \cdot b) \\ = (c, a) (d, b) \\ = φ (a, c) φ (b, d) \end{aligned}

To show that $φ$ is a bijection. I claim that $φ^{2}$ is it's inverse:

φ^{2} (a, b) = φ (b, a) = (a, b)

Thus $φ^{2} = I$ and therefore $φ$ is its own inverse, showing that $φ$ is a bijection.

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13

Question

Let $G$ and $H$ be groups and let $φ : G \to H$ be a Homomorphism (or Group Morphism). Prove that the image of $φ$ , namely $φ (G)$ , is a Subgroup of $H$ . Prove that if $φ$ is injective then $G ≅ φ (G)$ .

Proof

Using the Subgroup#The Subgroup Criterion conditions, we need to show that $\forall h, k \in φ (G) (h k, h^{- 1} \in φ (G))$ in order for $φ (G)$ to be a subgroup of $H$ (as already $φ (G) \subseteq H$ ).

Let $h, k \in φ (G)$ be arbitrary. Then notice that by $φ$ being a homomorphism. Then by being in the image then $\exists g_{h}, g_{k} \in G$ where $φ (g_{h}) = h$ and $φ (g_{k}) = k$ . Notice that by $φ$ being a homomorphism:

φ (g_{h} g_{k}) = φ (g_{h}) φ (g_{k}) = h k

So then we've determined that $h k \in φ (G)$ as desired. Similarly, since $h \in φ (G)$ then $\exists g_{h} \in G$ where $φ (g_{h}) = h$ . Then I claim that $h^{- 1} = φ (g_{h}^{- 1}) = φ (g_{h})^{- 1}$ . Verifying:

h h^{- 1} = φ (g_{h}) φ (g_{h}^{- 1}) = φ (g_{h} g_{h}^{- 1}) = φ (1) = e

And the other direction is checked similarly. Notice then that $h^{- 1} \in φ (G)$ as desired.

Therefore, we've shown that $φ (G) \leq H$ .

For the second part of the proof, suppose $φ$ is injective. We know that $φ$ is already a homomorphism, so if we can show that the restricted operator $φ^{⋆} : G \to φ (G)$ such that $φ^{⋆} : g \mapsto φ (g)$ is surjective, then we are done (since $φ^{⋆}$ is a restricted operator of an injective function, then it itself is already injective). But notice that if we consider any $h \in φ (G)$ then $\exists g \in G$ where $φ^{⋆} (g) = h$ , showing surjectivity! Thus then $G ≅ φ (G)$ .

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14

Question

Let $G, H$ be groups and let $φ : G \to H$ be a Homomorphism (or Group Morphism). Define the kernel of $φ$ to be ${g \in G | φ (g) = 1_{H}}$ . Prove that the kernel of $φ$ is a Subgroup of $G$ . Prove that $φ$ is injective iff the kernel of $φ$ is the identity subgroup of $G$ (the group with only the identity element).

Proof

To show it's a subgroup, we'll show closure of the restricted operator on $\ker (φ)$ for the binary operator, as well as inverses.

For the former, let $h, k \in \ker (φ)$ . Then $h, k \in G$ and $φ (h), φ (k) = 1_{H}$ . By $φ$ being homomorphic then:

φ (h k) = φ (h) φ (k) = 1_{H} 1_{H} = 1_{H}

Thus then since $h k \in G$ then $h k \in \ker (φ)$ , showing closure under the binary operation.

For the latter, let $h \in \ker (φ)$ , so like before then $φ (h) = 1_{H}$ where $h \in G$ . Then notice that $h^{- 1} \in G$ exists and:

φ (h^{- 1}) = φ (h)^{- 1} = 1_{H}^{- 1} = 1_{H}

thus then $h^{- 1} \in \ker (φ)$ .

$(\to)$ : Suppose $φ$ is injective. To show that $\ker (φ) = {1_{H}}$ , assume that it wasn't, that $\exists h \neq 1_{H} \in \ker (φ)$ . Then notice that both $h, 1_{H} \in \ker (φ)$ , so then $φ (h) = φ (1_{H}) = 1_{H}$ . But notice then that since $φ$ is injective that that implies that $h = 1_{H}$ , which is a contradiction! Thus then $\ker (φ) = {1_{H}}$ .

$(\leftarrow)$ : Suppose $\ker (φ) = {1_{H}}$ . Let $a, b \in \ker (φ)$ be arbitrary, and suppose $φ (a) = φ (b)$ . Clearly $a = b = 1_{H}$ so $φ$ is injective.

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16

Question

Let $A, B$ be groups and let $G$ be their direct product, $A \times B$ . Prove that the maps $π_{1} : G \to A$ and $π_{2} : G \to B$ defined by $π_{1} : (a, b) \mapsto a$ and $π_{2} : (a, b) \mapsto b$ are Homomorphism (or Group Morphism)s and find their Kernels.

Proof

Let $(a, b), (c, d) \in A \times B$ be arbitrary. Notice that:

\begin{aligned} π_{1} ((a, b) (c, d)) & = π_{1} ((a c, b d)) \\ = a c \\ = π_{1} (a, b) π_{1} (c, d) \end{aligned}

\begin{aligned} π_{2} ((a, b) (c, d)) & = π_{2} ((a c, b d)) \\ = b d \\ = π_{2} (a, b) π_{2} (c, d) \end{aligned}

thus both are homomorphisms. For their kernels we want to find ${g \in G | π_{1} (g) = 1_{A}}$ first. Namely ${(1_{A}, b) | b \in B}$ fits that bill, so then $\ker (π_{1}) = {(1_{A}, b) | b \in B}$ . A similar arguments shows that $\ker (π_{2}) = {(a, 1_{B}) | a \in A}$ .

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17

Question

Let $G$ be any Group. Prove that the map from $G$ to itself defined by $g \mapsto g^{- 1}$ is a Homomorphism (or Group Morphism) iff $G$ is an Abelian Group.

Proof

$(\Rightarrow)$ : Suppose $φ$ (defined above) is a homomorphism. Let $a, b \in G$ be arbitrary. Then:

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

$(\Leftarrow)$ : Suppose $G$ is abelian, and let $a, b \in G$ . Now notice:

\begin{aligned} φ (a b) & = (a b)^{- 1} \\ = b^{- 1} a^{- 1} \\ = φ (b) φ (a) \\ = φ (a) φ (b) & G is abelian \end{aligned}

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18

Question

Let $G$ be any Group. Prove that the map from $G$ to itself defined by $g \mapsto g^{2}$ is a homomorphism iff $G$ is an Abelian Group.

Proof

$(\Rightarrow)$ : Suppose $φ$ (defined above) is a homomorphism, and let $a, b \in G$ . Then:

\begin{aligned} φ (a b) & = φ (a) φ (b) \\ (a b)^{2} & = a^{2} b^{2} \\ a b a b & = a^{2} b^{2} \\ b a & = a b & (Left and Right Cancellation) \end{aligned}

$(\Leftarrow)$ : Suppose $G$ is an Abelian Group. Then:

\begin{aligned} φ (a b) & = (a b)^{2} \\ = a b a b \\ = a a b b \\ = a^{2} b^{2} \\ = φ (a) φ (b) \end{aligned}

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20

Question

Let $G$ be a group and let $Aut (G)$ be the set of all Isomorphisms from $G \to G$ . Prove that $Aut (G)$ is a Group under function composition (called the automorphism group of $G$ and the elements of $Aut (G)$ are called automorphisms of $G$ ).

Proof

First, let's establish that there's closure under these operations, which is easy to verify. If $φ_{1}, φ_{2} \in Aut (G)$ then $φ_{1} \circ φ_{2} \in Aut (G)$ since it still maps $G \to G$ , it's a composition of bijections (so $φ_{1} \circ φ_{2}$ is a bijection), and is a homomorphism if we let $g, j \in G$ and show that:

\begin{aligned} (φ_{1} \circ φ_{2}) (g j) & = φ_{1} (φ_{2} (g j)) \\ = φ_{1} (φ_{2} (g) φ_{2} (j)) \\ = φ_{1} (φ_{2} (g)) φ_{1} (φ_{2} (j)) \\ = (φ_{1} \circ φ_{2}) (g) (φ_{1} \circ φ_{2}) (j) \end{aligned}

For the individual properties:

By function composition applying from right to left, then $(φ_{1} φ_{2}) φ_{3} = φ_{1} (φ_{2} φ_{3})$ .
The identity element here is the identity map $φ_{0} : g \mapsto g$ . Namely, let $φ \in Aut (G)$ . Then:

(φ_{0} φ) = φ = (φ φ_{0})

For each $φ \in Aut (G)$ , by being an isomorphism and thus a bijection, then $φ^{- 1}$ is the inverse element. Namely:

φ φ^{- 1} = φ_{0} = φ^{- 1} φ

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