[2.4] Homomorphisms and Isomorphisms

Consider the subgroup $G_1 = \{\pm 1, \pm i\} \subset \mathbb{C}^*$. Where $\mathbb{C}^*$ is the group of non-zero complex numbers equipped with multiplication. Another way to write this is $G_1 = \{i^k \ | \ k = 0,1,2,3\}$. The multiplication table for this subgroup is

	$1$	$i$	$-1$	$-i$
$1$	$1$	$i$	$-1$	$-i$
$i$	$i$	$-1$	$-i$	$1$
$-1$	$-1$	$-i$	$1$	$i$
$-i$	$-i$	$1$	$i$	$-1$

And consider the cyclic subgroup $G_2$ of $S_4$, generated by $\rho$ where $1$ gets mapped to $2$, $2$ to $3$, $3$ to $4$ and $4$ to $1$. We can write this subgroup as

$$ \begin{align*} \rho = \begin{pmatrix}1 & 2 & 3 & 4 \end{pmatrix} \end{align*} $$

$$ \begin{align*} \langle \rho \rangle = \{\rho^n \ | \ n \in \mathbb{Z} \} \end{align*} $$

Note here that $\rho^2 \neq e$ and similarly, $\rho^3 \neq e$. In fact $\rho^4 = e$. Writing this as a multiplication table:

	$e$	$\rho$	$\rho^2$	$\rho^3$
$e$	$e$	$\rho$	$\rho^2$	$\rho^3$
$\rho$	$\rho$	$\rho^2$	$\rho^3$	$e$
$\rho^2$	$\rho^2$	$\rho^3$	$e$	$\rho$
$\rho^3$	$\rho^3$	$e$	$\rho$	$\rho^2$

We can see here that this multiplication table is just the first multiplication table relabeled. So $G_1$ and $G_2$ have the same multiplication table with a relabeling. Formally, (from 2.1)

The Preimage of a Function

Given some function $f: X \rightarrow Y$ and any subset $B \subseteq Y$, the preimage of $B$ in $X$ is $\{x \in X \ : \ f(x) \in B\}$. The conventional notation for the preimage of $B$ is $f^{-1}(B)$ even when the inverse of $f$ doesn’t exist. Based on this, we have the following proposition

Proposition 2.4.12

Let $\varphi : G \rightarrow H$ be a homomorphism of groups.

For each subgroup $A \subseteq G$, $\varphi(A)$ is a subgroup of $H$.
For each subgroup $B \in H$ $$ \begin{align*} \varphi^{-1}(B) &= \{ g \in G \ : \ \varphi(g) \in B\} \end{align*} $$ is a subgroup of $G$.

For $(b)$, $\varphi^{-1}(B)$ is the preimage of $B$. So we’re saying that the preimage of $B$ is a subgroup given that $B$ is a subgroup itself and $\varphi$ is a homomorphism.
Proof
For $(a)$, To show that $\varphi(A)$ is a subgroup of $H$, we need to show that $\varphi(A)$ is closed under multiplication and inverses. To see that it is closed under multiplication, we want to show for any $y_1, y_2 \in \varphi(A)$, $y_1y_2 \in \varphi(A)$. Since $y_1$ and $y_2$ are in $\varphi(A)$, then there must exist elements $x_1$ and $x_2$ such that $\varphi(x_1) = y_1$ and $\varphi(x_2) = y_2$.

$$ \begin{align*} y_1y_2 &= \varphi(x_1)\varphi(x_2) \\ &= \varphi(x_1x_2) \quad \text{(because $\varphi$ is a homomorphism)} \end{align*} $$

We know that $A$ is a subgroup of $G$ so $x_1x_2 \in A$. Therefore, $\varphi(x_1x_2)$ must also be in $\varphi(A)$ and therefore $y_1y_2 \in \varphi(A)$.
To see that $\varphi(A)$ is closed inverses, let $y \in \varphi(A)$, we want to show that $y^{-1} \in \varphi(A)$. Since $y \in \varphi(A)$, then there must exist an element $x \in A$ such that $y = \varphi(x)$. But $x$’s inverse must be in $A$ since $A$ is a subgroup so $x^{-1} \in A$. Furthermore, since $\varphi$ is a homomorphism, then $\varphi(x^{-1}) = (\varphi(x))^{-1} = y^{-1}$ by Proposition 2.4.11. Therefore $y^{-1} \in \varphi(A)$
For (b), Let $x_1, x_2 \in \varphi^{-1}(B)$, we want to show that $x_1x_2 \in \varphi^{-1}(B)$. Since $x_1, x_2 \in \varphi^{-1}(B)$, then they are elements of $G$ such that $\varphi(x_1)$ and $\varphi(x_2)$ are in $B$. But $B$ is a subgroup and so it is closed under multiplication. Therefore, $\varphi(x_1)\varphi(x_2) \in B$ …

	\(1\)	\(i\)	\(-1\)	\(-i\)
\(1\)	\(1\)	\(i\)	\(-1\)	\(-i\)
\(i\)	\(i\)	\(-1\)	\(-i\)	\(1\)
\(-1\)	\(-1\)	\(-i\)	\(1\)	\(i\)
\(-i\)	\(-i\)	\(1\)	\(i\)	\(-1\)

	\(e\)	\(\rho\)	\(\rho^2\)	\(\rho^3\)
\(e\)	\(e\)	\(\rho\)	\(\rho^2\)	\(\rho^3\)
\(\rho\)	\(\rho\)	\(\rho^2\)	\(\rho^3\)	\(e\)
\(\rho^2\)	\(\rho^2\)	\(\rho^3\)	\(e\)	\(\rho\)
\(\rho^3\)	\(\rho^3\)	\(e\)	\(\rho\)	\(\rho^2\)

The Preimage of a Function

References