Group Isomorphisms

The first isomorphism theorem

We begin with an easy lemma.

Lemma 1 For any homomorphism \(f:G\rightarrow G'\), \(\ker f\) is a normal subgroup of \(G\).

Proof

For any \(g\in G\) and \(x\in \ker f\),

\[f(gxg^{-1})=f(g)f(x)f(g^{-1})=f(g)e'f(g)^{-1}=f(g)f(g)^{-1}=e'.\]

Now, considering the equivalence relation defined by \(\ker f\),

\[x\sim y\iff xy^{-1}\in\ker f\]

we see from the following equation

\[f(y)=e'f(y)=f(xy^{-1})f(y)=f(xy^{-1}y)=f(x)\]

that \(x\sim y\iff f(x)=f(y)\). That is, \(\sim\) is nothing other than the equivalence relation defined by the function \(f\) ([Set Theory] §Examples of Equivalence Relations, ⁋Definition 2), and from the definition of a quotient group, the canonical map \(p:G\rightarrow G/\ker f\) is a homomorphism. Now, considering the canonical decomposition of \(f\), we obtain a bijection \(h:G/\ker f\rightarrow\im f\). Then for any \([x], [x']\in G/\ker f\), since

\[h([x][x'])=h([xx'])=f(xx')=f(x)f(x')=h([x])h([x'])\]

\(h\) is a homomorphism, and therefore an isomorphism.

Theorem 2 (The first isomorphism theorem) For any homomorphism \(f:G\rightarrow G'\), \(G/\ker f\cong \im f\) always holds.

On the other hand, using [Set Theory] §Examples of Equivalence Relations, ⁋Proposition 7, we obtain the following proposition.

Proposition 3 For any homomorphism \(f:G\rightarrow G'\) and any normal subgroup of \(G\), \(N\), there exists, satisfying \(f=\bar{f}\circ p\), a map \(\bar{f}:G/N\rightarrow G'\) if and only if \(N\leq \ker f\). In this case, since \(p\) is surjective, \(\bar{f}\) is uniquely determined.

The second isomorphism theorem

To prove the second isomorphism theorem, we must show the following lemma. In the following proposition, \(N\vee K\) denotes the smallest subgroup containing the union \(N\cup K\) in \(G\), that is, \(\langle N\cup K\rangle\), and \(NK\) denotes the set

\[NK=\{nk\mid n\in N,k\in K\}\]

Lemma 4 Let a group \(G\), a subgroup \(K\), and a normal subgroup \(N\) be given. Then the following hold.

1. \(N\cap K\) is a normal subgroup of \(K\).
2. \(N\) is a normal subgroup of \(N\vee K\).
3. \(NK=N\vee K=KN\) holds.

Proof

1. For any \(n\in N\cap K\) and \(k\in K\), since \(knk^{-1}\) is a product of elements of \(K\), it is an element of \(K\), and at the same time, since \(N\) is a normal subgroup of \(G\), it is an element of \(N\). Therefore, \(knk^{-1}\in N\cap K\).
2. It is clear that \(N\) is a subgroup of \(N\vee K\). Moreover, for any \(g\in N\vee K\) and \(n\in N\), we have \(gng^{-1}\in N\).
3. For any \(nk\in NK\), since \(n,k\in N\vee K\), we have \(nk\in N\vee K\). Thus it suffices to show only the reverse direction. Consider, for elements of \(N\) and \(K\), the subset containing all products \(n_1k_1\cdots n_rk_r\) in \(G\). One can easily verify that this set is a subgroup, and since this subgroup contains both \(N\) and \(K\), it also contains \(N\vee K\).¹
   Therefore, every element of \(N\vee K\) can be written in the form \(n_1k_1\cdots n_rk_r\). Now since \(N\) is a normal subgroup of \(N\vee K\), we have \(k_1n_2=n_2'k_1\) for some \(n_2'\in N\). By repeating this process, we can rewrite \(n_1k_1\cdots n_rk_r\) in the form of an element of \(NK\).

Theorem 5 (The second isomorphism theorem) Let a group \(G\), a subgroup \(K\), and a normal subgroup \(N\) be given. Then \(K/(N\cap K)\cong NK/N\) holds.

Proof

First, from the preceding lemma, \(N\) is a normal subgroup of \(NK=N\vee K=KN\). Meanwhile, since \(K\subseteq NK\), we can consider the composition of homomorphisms

\[K\overset{\iota}{\hookrightarrow}NK\overset{\pi}{\twoheadrightarrow}NK/N\]

Then, since

\[\ker(\pi\iota)=(\pi\iota)^{-1}(e)=\iota^{-1}(\ker\pi)=\iota^{-1}(N)=K\cap N\]

applying the first isomorphism theorem to \(\pi\iota\) yields

\[K/\ker(\pi\iota)=K/(K\cap N)\cong\im(\pi\iota)\]

However, every element of \(NK/N\) is of the form \(nkN\), and since there exists some \(n'\in N\) such that \(nk=kn'\), any element \(nkN\) of \(NK/N\) satisfies

\[nkN=kn'N=kN=\pi(k)=\pi(\iota(k))\in\im(\pi\iota)\]

which yields the desired result.

The third isomorphism theorem

Theorem 6 (The third isomorphism theorem) Let \(H\) and \(K\) be normal subgroups of a group \(G\), and let \(K\leq H\). Then \(H/K\) is a normal subgroup of \(G/K\) and \((G/K)/(H/K)\cong G/H\) holds.

Proof

The decomposition following [Set Theory] §Examples of Equivalence Relations, ⁋Definition 8.

The fourth isomorphism theorem

The following theorem is one of the most useful forms: given a group \(G\) and a normal subgroup of \(G\), \(N\), it shows that the function taking a subgroup containing \(N\) in \(G\), \(H\), to the subgroup of \(G/N\), \(H/N\), and conversely the function taking a subgroup of \(G/N\), \(\overline{H}\), to \(p^{-1}(\overline{H})\), are inverses of each other. The proof itself is a single line, but the point is that this bijection preserves intersections, indices, normality, and so on between subgroups; the proofs of these facts must each be given separately. Since these proofs are purely technical, we omit them.

Theorem 7 (The fourth isomorphism theorem) Let \(G\) be a group and let \(N\) be a normal subgroup of \(G\). Then there exists an inclusion-preserving bijection between the set of subgroups containing \(N\) in \(G\) and the set of subgroups of \(G/N\). Moreover, this bijection preserves all relations such as intersections, indices, and normal subgroups.

Coequalizer of homomorphisms

Now let two group homomorphisms \(f,g:G \rightarrow H\) be given. Earlier we saw that the equalizer of \(f,g\), \(\Eq(f,g)\), is always a subgroup of \(G\). Their coequalizer is somewhat more complicated.

First, considering the universal property of the coequalizer, \(q:H\rightarrow\CoEq(f,g)\) is initial among those satisfying \(q\circ f=q\circ g\). If we encountered such a situation in \(\Set\), then endowing \(H\) with the equivalence relation \(\sim\) generated by

\[f(x)\sim g(x)\qquad\text{for all $x\in G$}\]

and considering the projection \(H\rightarrow H/{\sim}\), this would be the coequalizer; but in \(\Grp\), we do not know whether the \(\sim\) defined above is compatible with the group operation of \(H\). That is, the following subset

\[S=\{f(x)g(x)^{-1}\mid x\in G\}\]

is not in general a normal subgroup, so \(H/S\) is not defined.

To resolve this, let \(\overline{S}\) be the normal closure of \(S\), that is, the smallest normal subgroup containing \(S\). Then the quotient by \(\overline{S}\) of \(H\), \(H/\overline{S}\), is well-defined.

Proposition 8 The quotient \(q: H \rightarrow H/\overline{S}\) defined as above is a coequalizer.

Proof

First, for any \(x\in G\), since \(f(x)g(x)^{-1}\in S\subseteq\overline{S}=\ker q\), we have \(q(f(x))=q(g(x))\), that is, \(q\circ f=q\circ g\) holds.

Suppose there exists a group homomorphism \(q': H \rightarrow H'\) satisfying \(q'\circ f=q'\circ g\). Then by Lemma 1, \(\ker q'\) is a normal subgroup, and by the condition \(q'\circ f=q'\circ g\), since

\[q'(f(x))=q'(g(x))\iff q'(f(x)g(x)^{-1})=e\]

we have \(f(x)g(x)^{-1}\in\ker q'\) for all \(x\in G\). Therefore, by the definition of \(\overline{S}\), we have \(\overline{S}\leq\ker q'\), and applying Proposition 3, we have \(q'=\overline{q'}\circ q\) for some \(\overline{q'}:H/\overline{S}\rightarrow H'\). Such \(\overline{q'}\) is unique because \(q\) is surjective.

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References

[Bou] Bourbaki, N. Algebra I. Elements of Mathematics. Springer. 1998.

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1. One can also easily show that the reverse inclusion holds. ↩