Isomorphism Theorems

The First Isomorphism Theorem

We begin with a simple 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 consider the equivalence relation defined by \(\ker f\)

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

From the equation

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

we see that \(x\sim y\iff f(x)=f(y)\). That is, \(\sim\) is nothing but the equivalence relation defined by the function \(f\) ([Set Theory] §Examples of Equivalence Relations, ⁋Definition 2), and from the definition of quotient groups, 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 \([x], [x']\in G/\ker f\),

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

so \(h\) 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 normal subgroup \(N\) of \(G\), there exists \(\bar{f}:G/N\rightarrow G'\) satisfying \(f=\bar{f}\circ p\) if and only if \(N\leq \ker f\).

The Second Isomorphism Theorem

To prove the second isomorphism theorem, we need the following lemma. In the following proposition, \(N\vee K\) denotes the smallest subgroup of \(G\) containing the union \(N\cup K\), i.e., \(\langle N\cup K\rangle\), and \(NK\) denotes the set

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

Lemma 4 Let \(K\) be a subgroup of a group \(G\) and \(N\) a normal subgroup of \(G\). 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 \(k\in K\), \(knk^{-1}\) is a product of elements of \(K\), so 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 \(N\vee K\). Also, for any \(g\in N\vee K\) and \(n\in N\), \(gng^{-1}\in N\) holds.
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 the reverse inclusion. Consider the subset of \(G\) containing all products \(n_1k_1\cdots n_rk_r\) of elements of \(N\) and \(K\). It is easy to 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 \(N\vee K\), there exists \(n_1'\in N\) such that \(k_1n_2=n_2'k_1\). 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 \(K\) be a subgroup of a group \(G\) and \(N\) a normal subgroup of \(G\). Then \(K/(N\cap K)\cong NK/N\) holds.

Proof

From the previous lemma, \(N\) is a normal subgroup of \(NK=N\vee K=KN\). On the other hand, since \(K\subset NK\), we can consider the composition of homomorphisms

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

Then

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

so applying the first isomorphism theorem to \(\pi\iota\),

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

is obtained. But every element of \(NK/N\) has the form \(nkN\), and for some \(n'\in N\), we can write \(nk=kn'\), so every element \(nkN\) of \(NK/N\) satisfies

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

and thus we obtain the desired result.

The Third Isomorphism Theorem

Theorem 6 (The Third Isomorphism Theorem) Let \(H\) and \(K\) be normal subgroups of a group \(G\) with \(K<H\). Then \(H/K\) is a normal subgroup of \(G/K\), and \((G/K)/(H/K)\cong G/H\) holds.

Proof

See the decomposition after [Set Theory] §Examples of Equivalence Relations, ⁋Definition 8.

The Fourth Isomorphism Theorem

The following theorem is used extensively, and while its proof is not difficult, there is too much to verify, so we omit the proof.

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

Coequalizer of Homomorphisms

Now suppose two group homomorphisms \(f,g:G \rightarrow H\) are given. We previously saw that the equalizer \(\Eq(f,g)\) of \(f\) and \(g\) is always a subgroup of \(G\). Their coequalizer is somewhat more complicated.

First, considering the universal property of coequalizers, \(q:H\rightarrow\CoEq(f,g)\) is initial among those satisfying \(q\circ f=q\circ g\). If we encountered this situation in \(\Set\), we would define an equivalence relation \(\sim\) on \(H\) as the relation generated by

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

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

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

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

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

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

Proof

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

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

so \(f(x)g(x)^{-1}\in\ker q'\) holds for all \(x\in g\). Therefore, by the definition of \(\overline{S}\), \(\overline{S}\leq\ker q'\), and applying Proposition 3 yields the desired result.

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References

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

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1. The reverse inclusion can also be easily shown. ↩