Group Theory Archives - Epsilonify

Prove that the group A4 is not abelian

The Mathematician — Sun, 30 Apr 2023 13:00:07 +0000

Prove that the group

A_4

is not abelian.

Proof. First of all, the alternating group of degree 4 is defined as:

\begin{align*} A_4 = \{1,(123),(124),(132),(134),(142),(143),(234),(243),(12)(34),(13)(24),(14)(23)\}. \end{align*}

To show that

A_4

is not abelian, we need to find a counterexample. That is, we need to find

\sigma, \tau \in A_4

such that

\sigma\tau \neq \tau\sigma

. So, let

\sigma = (123), \tau = (13)(24) \in A_4

. Then

\begin{align*} (123)(13)(24) = 1 \end{align*}

and

\begin{align*} (13)(24)(123) = (142). \end{align*}

But

(123)(13)(24) \neq (13)(24)(123)

, so

A_4

is not abelian.

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Every infinite cyclic group is isomorphic to the additive group of Z

The Mathematician — Wed, 26 Apr 2023 13:00:51 +0000

We will prove that every infinite cyclic group is isomorphic to the additive group of Z.

Define the following mapping:

\begin{align*} f: \mathbb{Z} &\longrightarrow \langle x \rangle \\ k &\longmapsto x^k, \end{align*}

where

\langle x \rangle

is the infinite cyclic group. Then this map is isomorphic.

Proof. To prove that the map

f

is isomorphic, we need to show that

f

it is well defined,
is injective,
is surjective,
and is a homomorphism.

Well defined. The map is already well defined since there is no ambiguity in the representations of elements in the domain. To be more clear with an example,

2 = \frac{2}{1} = \frac{4}{2}

etc, but

\frac{2}{1},\frac{4}{2} \not \in \mathbb{Z}

. So all elements in

\mathbb{Z}

are distinct. Same holds for

\langle x \rangle

since it is an infinite cyclic group.

Injectivity. All elements are distinct from what we have seen before, so if

x^a \neq x^b

, then

a \neq b

.

Surjectivity. The element

x^k

\langle x \rangle

is the image of

k

under

f

, so

f

is surjective.

Homomorphism. To prove that

f

is a homomorphism, we have done that here. Now we have shown that every infinite cyclic group is isomorphic to the additive group of Z.

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Any two cyclic groups of the same order are isomorphic

The Mathematician — Sat, 22 Apr 2023 13:00:14 +0000

Define the following map:

\begin{align*} f: \langle x \rangle &\longrightarrow \langle y \rangle \\ x^k &\longmapsto y^k \end{align*}

where

\langle x \rangle

and

\langle y \rangle

are cyclic groups of order

n \in \mathbb{Z}^{+}

. Then the map

f

is isomorphic.

Proof. We need to prove four things for the map

f

, i.e.:

it is well defined,
is injective,
is surjective,
and is a homomorphism.

The map $f$ is well defined. To show this, we need to prove that if

x^r = x^s

, then

f(x^r) = f(x^s)

. We know that

x^{r-s} = 1

, which implies that

n \mid r-s

, i.e.,

r = tn + s

for some

t \in \mathbb{Z}

. Now it is a straight calculation that:

\begin{align*} f(x^r) = f(x^{tn + s}) = y^{tn + s} = y^{tn}y^s = y^s = f(x^s). \end{align*}

So the map is well defined.

The map $f$ is injective. In order to show that it is injective, it needs to follow that if

f(x^r) = f(x^s)

, then

x^r = x^s

. Notice that

y^r = f(x^r) = f(x^s) = y^s

, and therefore,

y^{r-s} = 1

. We apply the same trick as we did when we proved that

f

is well defined map:

r = tn + s

for some

t \in \mathbb{Z}

. Therefore, we get:

\begin{align*} x^r = x^{tn + s} = x^{tn}x^s = x^s, \end{align*}

since both cyclic group have order

n

. This proves the injectivity.

The map $f$ is surjective. For each

y^k \in \langle y \rangle

, there exists an

x^k \in \langle x \rangle

such that

f(x^k) = y^k

. Since both groups do have order

n

, we have already proved the surjectivity.

The map $f$ is a homomorphism. This one is pretty straightforward. Take

a,b \in \mathbb{Z}

. Then

\begin{align*} f(x^{a} x^{b}) = f(x^{a + b}) = y^{a + b} = y^{a}y^{b} = f(x^a)f(x^b), \end{align*}

which proves the homomorphism. We finally proved that any two cyclic groups of the same order are isomorphic

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Prove that the symmetric group of degree at least 3 is non-abelian

The Mathematician — Tue, 18 Apr 2023 13:00:41 +0000

The symmetric group $S_n$ is non-abelian for $n \geq 3$ .

Proof. Since we need to look at degrees at least

3

, we can take

(12),(13) \in S_n

. Now note that:

\begin{align*} (12)(13) = (132) \end{align*}

and

\begin{align*} (13)(12) = (123). \end{align*}

In order to be abelian, it must be that

(12)(13) = (13)(12)

, but that is not the case since

\begin{align*} (12)(13) = (132) \neq (123) = (13)(12). \end{align*}

Therefore, the symmetric group

S_n

is non-abelian for

n \geq 3

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Is R a group under multiplication?

The Mathematician — Fri, 14 Apr 2023 13:00:08 +0000

Is $\mathbb{R}$ a group under multiplication?

Proof. We want to show that

(R, \times)

is not a group under multiplication. Let

r \in \mathbb{R}

.

Associativity is easily shown by the definition of

\mathbb{R}

. Further, we know that the identity element of

\mathbb{R}

1

since

r \times 1 = 1 \times r = r

.

Lastly, assuming by contradiction, we want to show that each element of

\mathbb{R}

has an inverse. All elements of

\mathbb{R}

do have an inverse, but zero doesn’t. In other words, there exist no

r \in \mathbb{R}

such that:

\begin{align*} 0 \times r = 1, \end{align*}

since

0

cancels everything out. Therefore,

\mathbb{R}

not a group under multiplication

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Prove that Z is not a group under multiplication

The Mathematician — Mon, 10 Apr 2023 13:00:41 +0000

Prove that $\mathbb{Z}$ is not a group under multiplication .

I do expect that when you are learning group theory and you search for an answer to the question above, you want to know how to tackle such a question. While this can be answered in one sentence, I will take you through the whole process of what you need to do.

Proof. Let

\mathbb{Z}

be the set of integers. To be a group under multiplication, i.e.,

(\mathbb{Z},\times)

, we must verify three things: associativity, identity, and inverse elements. Let

a,b,c \in \mathbb{Z}

.

It is straightforward by the definition of integers that

(a \times b)\times c = a \times (b \times c)

. So for all elements in

\mathbb{Z}

we know that the operation multiplication is associative.

Next, we need to find the identity element of

\mathbb{Z}

under multiplication. This is easy since

1

is identity element (note that

a \times 1 = 1 \times a = a

).

Finally, the last step is to find the inverse elements of

\mathbb{Z}

under multiplication. Now we take the element

a \in \mathbb{Z}

. Does this mean that there is an inverse element of

a

under multiplication? In other words, is there an element

a^{-1} \in \mathbb{Z}

such that:

\begin{align*} a \times a^{-1} = a^{-1} \times a = 1? \end{align*}

Assume it is true with

a \neq 1

. Then it means that

a^{-1} = \frac{1}{a} \in \mathbb{Z}

. But then

\frac{1}{a}

is a rational number, which isn’t contained in the set of integers. So

\mathbb{Z}

can’t be a group under multiplication since for all integers (expect for the identity element

1

) there is no inverse element.

A short counterexample you could give:

2

has no inverse element since

\frac{1}{2} \not \in \mathbb{Z}

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The image of a group homomorphism is a subgroup

The Mathematician — Sun, 05 Mar 2023 13:00:08 +0000

Let $G$ and $H$ be groups and let $\phi: G \longrightarrow H$ be a homomorphism. Then $\text{im}(\phi)$ is a subgroup of $H$ .

Proof. In order to show that

\text{im}(\phi)

is a subgroup of

H

, it must hold the subgroup criterion. Firstly, we know that

\text{im}(\phi)

is nonempty since

\phi(1_G) = 1_H

, where

1_G

and

1_H

are identities of

G

and

H

, respectively. The easiest way to see that is that

\phi(1_G) = \phi(1_G)\phi(1_G) = \phi(1_G)\phi(1_G)

, so it is getting cancelled. Secondly, let

x,y \in \text{im}(\phi)

. We need to show that

xy^{-1} \in \text{im}(\phi)

. Let

a,b \in G

such that

\phi(a) = x

and

\phi(b) = y

, and we know that

ab^{-1} \in G

. Then

\phi(b)^{-1} = y^{-1}

. So we get the following:

\begin{align*} xy^{-1} &= \phi(a)\phi(b)^{-1} \\ &= \phi(a)\phi(b^{-1}) \\ &= \phi(ab^{-1}). \end{align*}

Therefore, this means that

xy^{-1} \in \text{im}(\phi)

. So

\text{im}(\phi)

is a subgroup of

H

by the subgroup criterion.

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The kernel of a group homomorphism is a subgroup

The Mathematician — Wed, 01 Mar 2023 13:00:43 +0000

Let $G$ and $H$ be groups and let $\phi: G \longrightarrow H$ be a homomorphism. Then $\ker(\phi)$ is a subgroup of $G$ .

Proof. In order to show that

\ker(\phi)

is a subgroup of

H

, it must hold the subgroup criterion. For this proof, let

1_G

and

1_H

be the identities of

G

and

H

, respectively. Firstly, we know that

\ker(\phi)

is nonempty since it contains the identity element

1_G \in \ker(\phi)

. Secondly, let

x,y \in \ker(\phi)

. To satisfy the subgroup criterion, we only need to show that

xy^{-1} \in \ker(\phi)

. Since

\phi(x) = \phi(y) = 1_H

, we get the following:

\begin{align*} \phi(xy^{-1}) &= \phi(x)\phi(y^{-1}) \\ &= \phi(x)\phi(y)^{-1} \\ &= 1_H 1_H^{-1} \\ &= 1_H. \end{align*}

Therefore, this means that

xy^{-1} \in \ker(\phi)

. So

\ker(\phi)

is a subgroup of

G

by the subgroup criterion.

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Prove that the additive groups R and Q are not isomorphic

The Mathematician — Thu, 09 Feb 2023 13:00:39 +0000

Prove that the additive groups $\mathbb{R}$ and $\mathbb{Q}$ are not isomorphic

Proof. Assume that

\mathbb{R}

and

\mathbb{Q}

are isomorphic. Then there exists a mapping

\begin{align*} \phi: \mathbb{Q} \longrightarrow \mathbb{R} \end{align*}

which is bijective and is a group homomorphism. Let

x \in \mathbb{R}

such that

\phi(2) = x

and

q \in \mathbb{Q}

such that

\phi(q) = \sqrt{x}

. Then

\begin{align*} \phi(q)^2 = \phi(q^2) = x = \phi(2). \end{align*}

Since

\phi

is injective, we have that

q^2 = 2

. So this means that

q = \sqrt{2}

, which isn’t possible in

\mathbb{Q}

, a contradiction. Therefore, the additive groups

\mathbb{R}

and

\mathbb{Q}

are not isomorphic.

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Prove that the multiplicative groups R-{0} and C-{0} are not isomorphic

The Mathematician — Sun, 05 Feb 2023 13:00:11 +0000

Prove that the multiplicative groups $\mathbb{R}-{0}$ and $\mathbb{C}-{0}$ are not isomorphic

Proof. Assume that

\mathbb{R}-{0}

is isomorphic to

\mathbb{C}-{0}

. Then there exists a mapping

\begin{align*} \phi: \mathbb{C}-{0} \longrightarrow \mathbb{R}-{0} \end{align*}

which is bijective and is a group homomorphism. By the definition of a group homomorphism, we have that

\phi(1) = 1

. We can also rewrite that as:

\begin{align*} 1 = \phi(1) = \phi((-1)(-1)) = \phi(-1)\phi(-1) = \phi(-1)^2. \end{align*}

Since

\phi

is injective, it must hold that

\phi(-1) = -1

. Again, we can rewrite that as:

\begin{align*} -1 = \phi(-1) = \phi(i^2) = \phi(i)^2. \end{align*}

So this means that

\phi(i)^2 \in \mathbb{R}^{\times}

. But that is not possible since

\phi(i)^2

must be positive number in

\mathbb{R}^{\times}

, and therefore

\phi(i)^2 \not \in \mathbb{R}^{\times}

. A contradiction. So, the multiplicative groups

\mathbb{R}-{0}

and

\mathbb{C}-{0}

are not isomorphic.

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