Isomorphism theorems

Let $G_1$ and $G_2$ be two groups and let $f : G_1 \rightarrow G_2$ be a homomorphism of $G_1$ onto $G_2$. Furthermore, let $K = Ker(f)$. If $H_1 \leq G_1$, then (see Theorem 7.1.4) $f[H_1] \leq G_2$. If $H_2\leq G_2$ then it is readily seen that $H=f^{-1}[H_2]\leq G_1$: For if $h_1, h_2 \in H$, then

$$f(h_1h_2^{-1})=f(g_1)f(h_2)^{-1}\in H_2,$$

so $h_1h_2^{-1}\in H$ (see Chapter 2, exercise 2). Now since $e_2\in H_2$, where $e_2$ designates the identity of $G_2$, we have $K=f^{-1}(e_2)\subset H=f^{-1}(H_2)$ and so $f(H)=H_2$, since $f$ is onto and also using exercise 6(b) from Section 1.1. We therefore have shown that any subgroup $H_2$ of $G_2$ is of the form

$$H_2=f(H),$$ (8.7)

where $H \leq G$ satisfies the condition that $Ker(f) \subset H$.

Finally let $H$ be any subgroup of $G_1$ that contains the kernel, $K$. Then, of course,

$$H\subset H_1 = f^{-1}[f[H]].$$

(See exercise 6(a) of Section 1.1, where an equality is given if $f$ is 1-1, but the above inclusion holds for any $f$. Why?) However since $f$ is a homomorphism, we can show equality. For if $h_1 \in H_1$, then $f(h_1) \in f[H]$, so $f(h_1) = f(h)$, where $h\in H$. Thus $h_1 = hk$, where $k \in K$ (note $f(h^{-1}h_1) = e_2$), but $K \subset H$; whence $h_1 \in H$, and we have

$$f^{-1}[f[H]] = H.$$ (8.8)

With these results at our disposal, we are now in a position to prove the following result.

Theorem 8.3.1 (Correspondence Theorem)   Let $f : G_1 \rightarrow G_2$ be a homomorphism of the group $G_1$ onto the group $G_2$ with $K = Ker(f)$. Let $\{H_\alpha\}_{\alpha\in\Lambda}$ be the class of all subgroups of $G_1$ which contain $K$. The mapping (or correspondence)

$$\Phi:H_\alpha \rightarrow f[H_\alpha]$$

is a 1-1 correspondence between the family $\{H_\alpha\}_{\alpha\in\Lambda}$ and the class of all subgroups of $G_2$. Moreover, $H_\alpha \lhd G_1$ if and only if $f(H_\alpha)=f[H_\alpha]\lhd G_2$.

Remark 8.3.2   The condition that $f$ is onto in this theorem is really no restriction because if $f : G_1 \rightarrow G_2$ is not onto, we just replace $G_2$ with $f[G_1]$.

Proof: Part of the theorem has already been established in our remarks preceding the theorem. In particular, we have noted that the mapping $\Phi$ defined in the statement of the theorem is onto (see (8.7); i.e., $\Phi$ is the mapping of the family $\{H_\alpha\}_{\alpha\in\Lambda}$ to the family of all subgroups of $G_2$ given by

$$\Phi(H_\alpha) =f[H_\alpha].$$

We also note that each subgroup Hà is such that $K = Ker(f)\subset H_\alpha$, for all $\alpha\in\Lambda$. It is also clear that $\Phi$ is 1-1, for suppose $\Phi(H_\alpha) = \Phi(H_\beta)$, then $f[H_\alpha] = f[H_\beta]$. But from the remarks preceding the theorem in particular equation (8.8), we get

$$H_\alpha = f^{-1}[f[H_\alpha]]= f^{-1}[f[H_\beta]] =H_\beta.$$

Thus $\Phi$ is 1-1.

Finally, if $H_\alpha \subset G_1$ then since $f$ is onto, for arbitrary $g_2 \in G_2$ there exists a $g_1 \in G_1$ such that $g_2 = f(g_1)$. Thus

$$g_2f[H_\alpha] g_2^{-1} = f(g_1)f[H_\alpha] f(g_1)^{-1} = f[g_1 H_\alpha g_1^{-1}] = f[H_\alpha],$$

and so $f[H_\alpha]\subset G_2$. Conversely, if $f[H_\alpha]\subset G_2$, consider , where $g_1 \in G_1$. Then

$$f[g_1 H_\alpha g_1^{-1}] = f(g_1)f[H_\alpha] f(g_1)^{-1} = f[H_\alpha].$$

But then $\Phi(g_1 H_\alpha g_1^{-1}) =f[g_1 H_\alpha g_1^{-1}]= f[H_\alpha]=\Phi(H_\alpha)$ and since $\Phi$ is 1-1, we have $g_1 H_\alpha g_1^{-1}=H_\alpha$ and so $g_1 H_\alpha g_1^{-1}\lhd G_1$. We note that $g_1 H_\alpha g_1^{-1}$ is actually also one of the subgroups of $G_1$, which contains $K$ because $K$ is normal in $G_1$; for $K\subset H_\alpha$ implies $K=g_1 Kg_1^{-1} \subset g_1 H_\alpha g_1^{-1}$. $\Box$

We apply Theorem 8.3.1 to the particular case of the canonical homomorphism (see Example 7.1.2) $\kappa$ of a group $G$ onto a factor group $G/N$, where $N$ is, of course, a normal subgroup of $G$; i.e., $\kappa : G \rightarrow G/N$. We claim $Ker(\kappa ) = N$. Indeed, recall the identity in $G/N$ is $N$, so that $Ker(\kappa ) = \{x \in G \ \vert\ \kappa (x) = N\} = \{x \in G \ \vert\ xN = N\} = \{x\in G \ \vert\ x \in N\} = N$, as claimed. Thus, by Theorem 8.3.1, any subgroup of $G/N$ is of the form $\kappa [H]$ where $H \leq G$ and $N \subset H$. However,

$$\kappa [H] = \{hN \ \vert \ h \in H\} = H/N.$$ (8.9)

We have, therefore established the following result.

Corollary 8.3.3   Let $G$ be a group and let $N \lhd G$. Any subgroup of $G/N$ is of the form $H/N$ where $H$ is a subgroup of $G$ containing $N$. If $H_1$ and $H_2$ are two such subgroups of $G$, then $H_1 \not= H_2$ implies $H_1/N \not= H_2/N$. Moreover $H \lhd G$ if and only if $H/N \lhd G/N$.

We continue to assume $f : G_1 \rightarrow G_2$ is a homomorphism of the group $G_1$ onto $G_2$ with kernel $K$. Let $H$ be a normal subgroup of $G$ that contains $K$ and let $H_2=f[H]$. Consider the mappings

$$G_1 \stackrel{f}{\rightarrow} G_2 \stackrel{\kappa}{\rightarrow} G_2/H_2,$$

where $\kappa$ is the canonical map of $G_2$ onto $G_2/H_2$. Note $H_2\lhd G_2$, by Theorem 8.3.1, since $H \lhd G_1$.

The composite map

$$\kappa f : G_1\rightarrow G_2/H_2$$

is, of course, a homomorphism of $G_1$ onto $G_2/H_2$. Suppose $a\in G_1$ and $\kappa f(a)=H_2$, i.e., suppose $a \in Ker(\kappa f)$. Then $f(a)\in H_2$ and conversely. Hence, the $Ker(\kappa f)$ is (by equation (8.8)). Applying the FHT (Theorem 7.1.8), we have $G_2/H_2\cong G_1/H$, where the isomorphism of $G_1/H$ onto $G_2/H_2$ is given by $aH \longmapsto \kappa f(a)=f(a)H_2$. We summarize this in the following theorem, frequently called the first isomorphism theorem.

Theorem 8.3.4 (First Isomorphism Theorem)   Let $f : G_1 \rightarrow G_2$ be a homomorphism of the group $G_1$ onto the group $G_2$ with $Ker(f) = K$. Let $H \lhd G_1$ such that $K \subset H$. Then $f[H] \lhd G_2$, and

$$G_1/H\cong f[G_1]/f[H],$$

by the mapping $aH\longmapsto f(a)f[H]$.

Again, we consider the special case of a group $G$ and the canonical map $\kappa$ onto a factor group $G/N$. If $H \lhd G$ and $N \subset H$, then

$$\kappa [H] =H_2=H/N,$$

by equation (8.9). Thus Theorem 8.3.4 gives the following result.

Corollary 8.3.5   Let $G$ be a group and let $H$ and $N$ be normal subgroups of $G$ such that $N \subset H$. Then

$$G/H \cong (G/N)/(H/N).$$

Now assume that $H_1$ and $H_2$ are subgroups of a group $G$, and moreover that $H_2 \lhd G$. Then, in particular, $h_1H_2 = H_2h_1$ for all $h_1 \in H_1$, and so clearly $H_1H_2 = H_2H_1$. This implies by Theorem 4.2.1, that $H_1H_2 \leq G$. As we will have occasion to use this fact in the future, we state it here as the following result.

Proposition 8.3.6   If $H_1\leq G$, $H_2 \leq G$ and $H_2 \lhd G$, then $H_1H_2 \leq G$.

(Of course $H_2 \subset H_1H_2$ and since $H_2 \lhd G$, we have $H_2 \leq H_1H_2$.) Next consider the mapping

$$\phi : H_1 \rightarrow H_1H_2/H_2$$

given by $\phi(h_1) = h_1H_2$, where $h_1 \in H_1$. This map is a homomorphism of $H_1$ into $H_1H_2/H_2$ (see exercise 1 for this section). We claim tht $\phi$ is, in addition, onto $H_1H_2/H_2$. Indeed, for any coset of $H_2$ in $H_1H_2$ is of the form $h_1h_2H_2 = h_1H_2$, where $h_1 \in H_1$ and $h_2 \in H_2$. The kernel, $Ker(\phi)$, consists of those $h_1 \in H_1$ such that $\phi(h_1) = h_1H_2 = H_2$, i.e., those elements of $H_1 \cap H_2$ or $Ker(\phi) = H_1 \cap H_2$. Thus applying the FHT (Theorem 7.1.8) to $\phi$ yields the second fundamental isomorphism theorem.

Theorem 8.3.7 (Second Isomorphism Theorem)   If $H_1$ and $H_2$ are subgroups of a group $G$ and $H_2$ is also normal in $G$, then $H_1 \cap H_2 \lhd H_1$ and

$$H_1H_2/H_2 \cong H_1/H_1 \cap H_2.$$

The isomorphism is given by the mapping $h_1(H_1 \cap H_2) \cong h_1H_2$, where $h_1 \in H_1$.

The second isomorphism theorem can probably best be remembered by the following mnemonic device:

Label the vertices of the figure as indicated, it being immaterial which side are writes $H_1$ or $H_2$ on. One then reads the isomorphism by reading ``modulo the opposite sides.'' Should $H_1$ also be normal in $G$, then we obtain, symmetrically, $H_1H_2/H_1 \cong H_2/(H_1 \cap H_2)$, which may be read by reading ``modulo'' the other pair of opposite sides of the figure.

There is another fundamental theorem of isomorphism (the third isomorphism theorem) due to Zassenhaus, but we postpone a consideration of this theorem until we reach the section to which it is most relevant.