Group Homomorphisms

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

We begin with an easy lemma.

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

Proof

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

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

Now consider the equivalence relation defined by \(\ker f\):

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

From the identity

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

we see that \(x\sim y\iff f(x)=f(y)\). That is, \(\sim\) is nothing other than the equivalence relation induced by the function \(f\) ([Set Theory] §Examples of Equivalence Relations, ⁋Definition 2), and by the definition of a quotient group the canonical map \(p:G\rightarrow G/\ker f\) is a homomorphism. 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 \(f:G\rightarrow G'\), we always have \(G/\ker f\cong \im f\).

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 \(N\) of \(G\), there exists \(\bar{f}:G/N\rightarrow G'\) satisfying \(f=\bar{f}\circ p\) if and only if \(N\leq \ker f\). In this case, since \(p\) is surjective, \(\bar{f}\) is uniquely determined.

The second isomorphism theorem

To prove the second isomorphism theorem, we need the following lemma. In the next proposition, \(N\vee K\) denotes the smallest subgroup of \(G\) containing the union \(N\cup K\), that is, \(\langle N\cup K\rangle\), 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 let \(N\) be 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. \(NK=N\vee K=KN\).

Proof

1. For any \(n\in N\cap K\) and \(k\in K\), the element \(knk^{-1}\) is a product of elements of \(K\), hence lies in \(K\), and at the same time, since \(N\) is a normal subgroup of \(G\), it also lies in \(N\). Therefore \(knk^{-1}\in N\cap K\).
2. That \(N\) is a subgroup of \(N\vee K\) is obvious. 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 the reverse inclusion. Consider the subset of \(G\) consisting of all products of the form \(n_1k_1\cdots n_rk_r\) with \(n_i\in N\) and \(k_i\in K\). One easily checks that this subset is a subgroup, and since it contains both \(N\) and \(K\), it also contains \(N\vee K\).¹
   Hence 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 \(N\vee K\), for \(k_1n_2\) there exists \(n_2'\in N\) 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 let \(N\) be a normal subgroup of \(G\). Then \(K/(N\cap K)\cong NK/N\).

Proof

First, from the preceding lemma, \(N\) is a normal subgroup of \(NK=N\vee K=KN\). On the other hand, since \(K\subseteq NK\), we may 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 \(\pi\iota\) yields

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

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

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

which gives 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 \(K\leq H\). Then \(H/K\) is a normal subgroup of \(G/K\) and \((G/K)/(H/K)\cong G/H\).

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 \(N\) of \(G\), it shows that the function taking a subgroup \(H\) of \(G\) containing \(N\) to the subgroup \(H/N\) of \(G/N\), and conversely the function taking a subgroup \(\overline{H}\) of \(G/N\) 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; the proofs of these facts must each be given separately. These proofs are purely technical, so 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 of \(G\) containing \(N\) and the set of subgroups of \(G/N\). Moreover, this bijection preserves all relations such as intersection, index, and normal subgroup.

Coequalizer of group homomorphisms

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

First, from 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 \(\Set\), we would define an equivalence relation \(\sim\) on \(H\) by 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; but in \(\Grp\) we do not know whether the \(\sim\) defined above is compatible with the group operation of \(H\). That is, the 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 \(H/\overline{S}\) 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

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

Suppose a group homomorphism \(q': H \rightarrow H'\) satisfies \(q'\circ f=q'\circ g\). Then by Lemma 1, \(\ker q'\) is a normal subgroup, and from the condition \(q'\circ f=q'\circ g\) we have

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

so \(f(x)g(x)^{-1}\in\ker q'\) holds for all \(x\in G\). Therefore, by the definition of \(\overline{S}\), we have \(\overline{S}\leq\ker q'\), and applying Proposition 3 yields a homomorphism \(\overline{q'}:H/\overline{S}\rightarrow H'\) satisfying \(q'=\overline{q'}\circ q\). 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. The reverse inclusion can also be easily verified. ↩