§3 Maps of Groups

For all $t_{1}$, $t_{2}\in(\mathbb{R},+)$,

For all $M_1,M_2\in GL_n(\mathbb{R})$,

For all $t_1,t_2\in(\mathbb{R},+)$,

(a) A vector space $V$ is a group because $V$ satisfies

(1) Associativity: addition in $V$ is associative, i.e. for $u,v,w\in V$,

(2) Identity: there exists an element $0\in V$ such that for every $v\in V$,

(3) Inverses: for every $v\in V$, there exists $-v\in V$ such that

(b) A linear map $\phi:V\to W$ between any two vector spaces $V$ and $W$ is a group homomorphism because $\phi$ satisfies

(1) The homomorphism property: for all $u,v\in V$,

(2) Preservation of the identity: since $\phi$ is linear,

where $0_V$ and $0_W$ are the identity elements of $V$ and $W$, respectively;

(3) Preservation of inverses: for every $v\in V$,

This follows from the linearity of $\phi$, which involves scalar multiplication and additive inverses.

(c) Any linear subspace is a subgroup because

• a linear subspace $U\subseteq V$ is also a group under addition, since it inherits the group structure of $V$;

• the identity $0_V$ of $V$ is also the identity in $U$;

• inverses in $U$ are inherited from $V$.

(a) For any $g\in G$,

Let $h$ be the inverse of $\phi(g)$. Then

(b)

Consider the identity map $\mathrm{id}_G:G\to G$ defined by $\mathrm{id}_G(g)=g$ for every $g\in G$. To show that $\mathrm{id}_G$ is a homomorphism, we need to show that for all $g_1,g_2\in G$,

For any $g_1,g_2\in G$,

and

Therefore,

Hence $\mathrm{id}_G:G\to G$ is a homomorphism.

To prove that the composition of two homomorphisms $\phi$ and $\psi$ is again a homomorphism, we need to verify that for all $g_1,g_2\in G$,

By the definition of composition,

Using the homomorphism property of $\phi$,

Then using the homomorphism property of $\psi$,

By the definition of composition,

Thus,

Therefore, if $G\stackrel{\phi}{\longrightarrow}H$ and $H\stackrel{\psi}{\longrightarrow}K$ are homomorphisms, then $\psi\circ\phi$ is a homomorphism.

To prove that the inclusion map $i:H\hookrightarrow G$ is a homomorphism, where $H\subset G$ is a subgroup, we need to verify that for all $h_1,h_2\in H$,

The inclusion map $i:H\hookrightarrow G$ is defined by

for every $h\in H$.

Thus, for any $h_1,h_2\in H$,

Therefore,

At the same time, $h_1\cdot h_2\in H$, and by the definition of the inclusion map,

Hence, for all $h_1,h_2\in H$,

Therefore, the inclusion map $i:H\hookrightarrow G$ is a homomorphism.

$\ker(\phi)$ is a subgroup of $G$ because

(1) If $\phi(g_1)=\phi(g_2)=1_H$, then

(2) $\phi(1_G)=1_H$, so $1_G\in\ker(\phi)$.

(3) If $\phi(g)=1_H$, then

Similarly, $\mathrm{im}(\phi)$ is a subgroup of $H$ because

(1) $h_i=\phi(g_i)$ implies

(2) $\phi(1_G)=1_H$, so $1_H\in\mathrm{im}(\phi)$.

(3) $h=\phi(g)$ implies

To prove that $\exp:(\mathbb{R},+)\to(\mathbb{R}_{>0},\times)$ is a group isomorphism, we need to show two things:

(1) Homomorphism: $\exp$ preserves the group operation, i.e. for all $x,y\in\mathbb{R}$,

(2) Bijection: $\exp$ is bijective, meaning that it is both injective and surjective.

Let us verify these.

For $x,y\in\mathbb{R}$,

Using the exponential law,

Therefore,

so $\exp$ is a homomorphism.

To show that $\exp$ is injective, suppose that for some $x,y\in\mathbb{R}$,

Then

Taking the natural logarithm of both sides (which is valid since $e^x>0$ for all $x$),

Therefore, $\exp$ is injective.

To show that $\exp$ is surjective, we must show that for every $y\in\mathbb{R}_{>0}$, there exists an $x\in\mathbb{R}$ such that $\exp(x)=y$.

Since $y>0$, we may take $x=\ln(y)\in\mathbb{R}$, where $\ln$ denotes the natural logarithm. Then

Therefore, $\exp$ is surjective.

Since $\exp$ is both a homomorphism and a bijection, it is a group isomorphism between $(\mathbb{R},+)$ and $(\mathbb{R}_{>0},\times)$.

Hence,

is a group isomorphism.

Let $G=\langle g\rangle$ be a cyclic group of order $n$. Define $\phi:G\to\mathbb{Z}/n\mathbb{Z}$ by

(1) Homomorphism:

On the other hand,

The two sides are equal, so $\phi$ is a homomorphism.

(2) Injectivity: If $\phi(g^a)=\phi(g^b)$, then $a\equiv b\pmod n$, which implies $g^a=g^b$.

(3) Surjectivity: For every $k\in\mathbb{Z}/n\mathbb{Z}$, we have

Clearly, $\phi^{-1}$ is a bijection. It remains to prove that $\phi^{-1}$ is a homomorphism.

Let $g_1,g_2\in G$, and let $h_1=\phi(g_1)$, $h_2=\phi(g_2)$. Then

(1) Associativity follows directly from associativity in $G_1$ and $G_2$.

(2) Identity:

is the identity element.

(3) Inverse:

$\mathrm{Aut}(X)$ is a group under composition because

(1) composition of functions is associative:

$\begin{aligned} (2)~\mathrm{id}_X:X&\to X \text{ is the identity, since } f\circ\mathrm{id}_X=\mathrm{id}_X\circ f=f.\ x&\mapsto x \end{aligned}$

(3) $f^{-1}$ is the inverse of $f$:

Let $G$ be an arbitrary group. Our goal is to construct an injective group homomorphism

thereby proving that $G$ is isomorphic to a subgroup of $\mathrm{Sym}(G)$.

For each element $g\in G$, define a function $L_g:G\to G$ by left multiplication:

Since $G$ is a group, every $L_g$ is a bijection, with inverse $L_{g^{-1}}$. Hence

Define

by

• Homomorphism property: For all $g_1,g_2\in G$ and $h\in G$,

Therefore,

so $\phi$ is a group homomorphism.

• Injectivity: Suppose $\phi(g)=\phi(h)$ for some $g,h\in G$. Then for every $k\in G$,

In particular, taking $k$ to be the identity element $e$ of $G$,

Therefore, $\phi$ is injective.

Since $\phi$ is an injective homomorphism, $G$ is isomorphic to the subgroup $\phi(G)$ of $\mathrm{Sym}(G)$.