Definition and Elementary Properties

We return to a consideration of mappings from one group to another.

Definition 7.1.1   Let $G_1$ and $G_2$ be two groups (both denoted multiplicatively) and let $f : G_1 \rightarrow G_2$. The mapping $f$ is called a homomorphism if $f(ab) = f(a)f(b)$ for all $a,b \in G_1$. If $f$ is an onto homomorphism, i.e., $G_2 = f[G_1]$, then $G_2$ is called a homomorphic image of $G_1$.

In words, a homomorphism is just a map from one group to another which preserves the operation. Let us consider two examples of homomorphisms: the first rather general, the second quite specific.

Example 7.1.2   Let $G$ be a group and $N \lhd G$. Consider the canonical (or natural) mapping $\kappa$ of $G$ onto $G/N$ given by $\kappa (a) = aN$ for all $a \in G$. $\kappa$ is clearly onto $G/N$, and

$$\kappa (ab) = abN = (aN) (bN) = \kappa(a)\kappa (b).$$

Thus $G/N$ is a homomorphic image of $G$. This example shows that a factor group of a group is always a homomorphic image of the group. $\Box$

Example 7.1.3   Let $G =\mathbb{C}^n$, as in Example 2.1.14. Map $\mathbb{C}^n$ to itself by

$$f : \mathbb{C}^n \rightarrow \mathbb{C}^n$$

where $f(\alpha_1, ..., \alpha_n) = (\alpha_1, 0, ..., 0)$. It is easy to see that $f$ is a homomorphism of $\mathbb{C}^n$ into itself (see exercise 1 for this section). $\Box$

We now consider some general properties of homomorphisms.

Theorem 7.1.4   Let $f : G_1 \rightarrow G_2$ be a homomorphism. Then $f[G_1] \leq G_2$.

Proof: Let $e$ be the identity of $G_1$. We have $f(e) \in f[G_1]$, so $f[G_1] \not=\emptyset$. Let $f(a), f(b) \in f[G_1]$. Then $f(a)f(b) = f(ab) \in f[G_1]$. But $f(e) = f(e\cdot e)= f(e)f(e)$ which implies by cancellation that $f(e)$ is the identity element of $G_2$. Finally, $f(a)f(a^{-1}) = f(aa^{-1}) =f(e)$, so $f(a)^{- 1} = f(a^{-1}) \in f[G_1]$. This shows that $f[G_1] \leq G_2$ by Definition 2.2.3. $\Box$

Let us remark that in the above proof we showed two other important properties of a homomorphism. They are:

• $f(e_1)= e_2$, where $e_i$ is the identity of $G_i$ ($i = 1, 2$),

• $f(a^{-1}) = f(a)^{-1}$, for every $a\in G_1$.

Definition 7.1.5   Let $f : G_1 \rightarrow G_2$ be a homomorphism, and let $e$ designate the identity of $G_1$ as well as of $G_2$. Let $K = \{a \in G_1 \ \vert\ f(a) = e\}$. The set $K$ is called the kernel of $f$. We write $Ker(f)$.

We shall presently show that if $f : G_1 \rightarrow G_2$ is a homomorphism then $Ker(f) \lhd G_1$, but first we note the following basic property.

Theorem 7.1.6   $f$ is 1-1 if and only if $Ker(f) = \{e\}$.

Proof: If $f$ is 1-1, since $f(e) = e$, $e$ can be the only element which maps to $e$, i.e., $Ker(f) = \{e\}$. Conversely, if $Ker(f) = \{e\}$, then if $f(a) = f(b)$, we have $f(a^{-1}b) = f(a^{-1})f(b) = f(a)^{- 1}f(b) = e$ since $f(a) = f(b)$. Thus $a^{-1}b \in K = \{e\}$. So $a^{-1}b = e$, and $a = b$. $\Box$

Theorem 7.1.7   If $f : G_1 \rightarrow G_2$ is a homomorphism then $Ker(f) \lhd G_1$.

Proof: We have already seen that $e \in Ker(f)$. If $a,b \in Ker(f)$, then $f(a) = e$ and $f(b) = e$, so $f(ab) = f(a)f(b)= e$; hence $ab \in Ker(f)$. Also $f(a^{-1}) = f(a)^{-1} = e^{-1} = e$, i.e., if $a \in Ker(f)$, then $a^{-1} \in Ker(f)$. We have so far shown that $Ker(f) \leq G_1$. To see that it is normal, suppose $k \in Ker(f)$ and $a \in G$. Then $f(aka^{-1}) = f(a)f(k)f(a^{-1}) = f(a)ef(a)^{-1} = e$. Therefore $k \in Ker(f)$ implies that $aka^{-1} \in Ker(f)$. This says $aKer(f)a^{-1} \subset Ker(f)$ for all $a\in G_1$. This is sufficient to show $Ker(f) \lhd G$, according to Proposition 6.1.4. $\Box$

In the case of the canonical map $\kappa : G \rightarrow G/N$ (see Example 7.1.2), $\kappa (a) = N$ (the identity element of $G/N$) if and only if $aN = N$ if and only if $a \in N$. Thus $Ker(\kappa ) = N$.

We observe (see exercise 2 for this section) that if $G_1, G_2, G_3$ are groups and if $f_1 : G_1 \rightarrow G_2$ and $f_2 : G_2 \rightarrow G_3$ are homomorphisms then $f_2f_1 : G_1 \rightarrow G_3$ is a homomorphism.

Suppose again that $f : G_1 \rightarrow G_2$ is a homomorphism with $K = Ker(f)$. We observed, for general mappings in section 1.1 that there is associated with $f$ a factorization

$$G_1 \stackrel{\kappa}{\rightarrow} \overline{G_1} \stackrel{g}{\rightarrow} f[G_1]\stackrel{i}{\rightarrow} G_2,$$

such that $f = ig\kappa$, $\kappa$ is onto, $g$ is 1-1 and onto, while $i$ is an injection mapping. Also recall that was a set of equivalence classes, determined by the equivalence relation: $a \sim b$ if and only if $f(a) = f(b)$. Consider the equivalence class of $a\in G_1$, namely $[a]$. Now $b\in [a]$ if and only if $b \sim a$ if and only if $f(a) = f(b)$ if and only if $f(a^{-1}b) = e$ if and only if $a^{-1}b \in K$ if and only if $b \in aK$. Thus $[a] = aK$ and $\overline{G_1}$ is precisely $G_1/K$ and $\kappa$ is the canonical homomorphism. Thus we have

$$G_1 \stackrel{\kappa}{\rightarrow} G_1/K \stackrel{g}{\rightarrow} f[G_1] \stackrel{i}{\rightarrow} G_2.$$

Now $i$, being an injection mapping, is an isomorphism of $f[G_1]$ into $G_2$. Finally we claim that $g$ is an isomorphism of $G_1/K$ onto $f[G_1]$. We know, in general that $g$ is 1-1 and onto, thus all that needs to be shown is that $g$ preserves the operation. Now $g(aK) = f(a)$, recalling the definition of $g$, so

$$g(aKbK)=g(abK)=f(ab)=f(a)f(b)=g(aK)g(bK).$$

Consequently, we have $f[G_1] \cong G_1/Ker(f)$.

We have thus established the fundamental result stated below.

Theorem 7.1.8 (Fundamental Homomorphism Theorem (FHT)   )

(I)

If $N \lhd G$, a group, then $G/N$ is a homomorphic image of $G$.

(II)

If $f : G_1 \rightarrow G_2$ is a homomorphism, then $G_1/Ker(f) \cong f[G_1]$. In particular, if $G_2$ is a homomorphic image of $G_1$, then $G_1/Ker(f) \cong G_2$.

We note that the significance of this theorem is that it relates two seemingly unrelated concepts, i.e., concepts of factor group and homomorphic image. In particular, the FHT basically says that these two concepts coincide.

Example 7.1.9   Let $f : \mathbb{Z}_6\rightarrow \mathbb{Z}_3$ defined by $\left( \begin{array}{cccccc} 0 & 1 & 2 & 3 & 4 & 5 \\ 0 & 1 & 2 & 0 & 1 & 2 \end{array} \right)$. It is not hard to verify that $f$ is a homomorphism of $\mathbb{Z}_6$ onto $\mathbb{Z}_3$. $Ker(f) = \{0,[3]\}$ is the subgroup of $\mathbb{Z}_6$ generated by $[3]$, i.e., $\langle [3] \rangle$. (Here $[a]$ is in the notation of Exercise # 4 in §4.3.) FHT implies $\mathbb{Z}_6/\langle [3] \rangle \cong \mathbb{Z}_3$. We note that this example could also have been given as follows: Let $f : \mathbb{Z}/6\mathbb{Z}\rightarrow \mathbb{Z}/3\mathbb{Z}$ defined by $f(i + 6\mathbb{Z}) = i + 3\mathbb{Z}$ for any $i \in \mathbb{Z}$. The reader should verify again that this is a homomorphism, find its kernel and state the conclusion of FHT in this case. $\Box$