The Isomorphism Theorems

Theorem (First isomorphism theorem)

Let \(G, G'\) be groups, let \(f:G \to G'\) be a homomorphism and let \(H = \operatorname{Ker}(f)\). Then \(G / H \cong \operatorname{Im}(f)\).

Proof: Since \(H \triangleleft G\), \(G / H\) is a group. Let \(f':G / H \to G'\) be such that \(aH \mapsto f(a)\). We shall prove that \(f'\) is well defined, i.e., that it sends members of the same equivalence class to the same images. Given \(a,b \in G\), suppose that \(aH = bH\), i.e., that \(b^{-1}a \in H\). Thus,

\[f(b^{-1}a)=1_{G'} \implies (f(b))^{-1} f(a) = 1_{G'}\implies f(b) = f(a) \implies f'(aH) = f'(bH).\]

We shall now prove that \(f'\) is a homomorphism. Given \(aH, bH \in G / H\), note that

\[f'(aH \cdot bH) = f'((a \ast b)H)= f(a \ast b) =f(a) \ast_{G'}f(b) = f'(aH)\ast_{G'} f'(bH).\]

Thus, \(f\) is a homomorphism. For injectivity, let \(aH \in \operatorname{Ker}(f')\). Hence,

\[f'(aH) = 1_{G'} \implies f(a) = 1_{G'} \implies a \in \operatorname{Ker}(f) = H \implies aH=H.\]

Finally, we shall prove that \(f'\) is surjective, i.e., that \(\operatorname{Im}(f') = \operatorname{Im}(f)\). Note that

\[b \in \operatorname{Im}(f') \iff \exists a \in G (f'(aH)=b) \iff \exists a \in G (f(a)=b) \iff b \in \operatorname{Im}(f).\]

Thus, \(f'\) is an isomorphism, from which we conclude the result.

Theorem (Third isomorphism theorem)

Let \(K \leq H \leq G\) be groups such that \(K \triangleleft G\) and \(H \triangleleft G\). Then \(H / K \triangleleft G / K\) and \(G / H \cong (G/K) / (H / K)\).

Proof: We shall first prove that \(H / K \triangleleft G / K\). Let \(aK \in G / K\). We must show that \((aK) \cdot H /K \cdot (aK)^{-1} \subseteq H / K\). Let \(bK \in H / K\), hence

\[(aK) \cdot (bK) \cdot (aK)^{-1} = (aba^{-1})K \in H / K.\]

We conclude that \(H / K \triangleleft G / K\). Now, define \(f:G/K \to G / H\) such that \(aK \mapsto aH\). We shall prove that \(f\) is well defined. Take \(a,b \in G\) such that \(aK = bK\). Then,

\[b^{-1}a \in K \subseteq H \iff b^{-1} a \in H \iff aH = bH \implies f(aK) = f(bK).\]

Let us prove that \(f\) is a homomorphism:

\[f(aK \cdot bK) = f(a \cdot b \cdot K) \quad= ab H = aH \cdot bH =f(aK) \cdot f(bK),\]

from which we conclude that \(f\) is a homomorphism. Finally, note that

\[aK \in\operatorname{Ker}(f) \iff f(aK) =H \iff aH =H \iff a \in H \iff aK \in H / K,\]

so that \(\operatorname{Ker}(f) = H / K\). Finally, applying the first isomorphism theorem, we conclude the result.

Theorem (commutativity of normal subgroups)

Let \(N,H \leq G\) with \(N \triangleleft G\). Then \(NH=HN\).

Proof: Exercise

Theorem (Second isomorphism theorem)

Let \((G, \cdot)\) be a group, \(H \leq G\), \(N \triangleleft G\). Then \(N \cap H \triangleleft H\), \(NH \leq G\) and \(H / (N \cap H) \cong NH / N\).

Proof: We know that \(N \cap H \subseteq H\), \(N \cap H \subseteq N\). Let \(x \in H\). Then \(x \in G\). We must show that \(x(N \cap H)x ^{-1} \subseteq N \cap H\). Clearly, \(x(N \cap H)x ^{-1} \subseteq H\). Since \(N \triangleleft G\), we have \(x(N \cap H)x ^{-1} \subseteq x N x ^{-1} \subseteq N\). Thus, \(x(N \cap H)x ^{-1} \subseteq N \cap H\) and we have proved the normality of \(N \cap H\). The proof that \(NH\leq G\) is left as an exercise. Finally, define \(f:H \to G / N\) such that \(h \mapsto h N\). Note that \(f\) is a homomorphism, since \(f(g \cdot h) = g h N = (gN)(hN) = f(g) \cdot f(h)\). Moreover,

\[h \in \operatorname{Ker}(f) \iff f(h) = N \iff hN =N \iff h \in N,\]

so that \(\operatorname{Ker}(f) = N \cap H\). Finally, we prove that \(\operatorname{Im}(f) = NH /N\). Let \(gN \in \operatorname{Im}(f)\). Hence, there exists \(h \in H\) such that \(hN = gN\). Hence, \(g \in HN = NH\), so \(gN \in NH / N\). Now, let \(g \in HN / N\). Then, there exists \(x \in HN\) such that \(g = xN \in HN / N\). Hence, \(x = h_{1} n_{1}\), with \(h_{1} \in H\), \(n_{1} \in N\). Thus,

\[g = xN = (h_{1} \cdot n_{1})N = h_{1} N,\]

so \(g = f(h_{1}) \in \operatorname{Im}(f)\). The result follows by applying the first isomorphism theorem.

Exercise (Inverse image of a subgroup)

If \(f:G \to G^{\ast}\) is a homomorphism and \(S^{\ast}\leq G^{\ast}\). Then \(f^{-1}(S^{\ast}) = \{ x \in G : f(x)\in S^{\ast} \}\)

Theorem (Correspondence)

Let \((G, \cdot)\) be a group, \(H \triangleleft G\) and \(p:G \to G / H\) such that \(p(g)=gH\). Then \(K \mapsto p(K) = K / H\) is a bijection between the subgroups of \(G\) that contain \(H\) and the subgroups of \(G/H\). Moreover, if \(K\) is such that \(H\leq K\leq G\) and \(K^{\ast}:=K / H\), we have that

\(L\leq K \iff L^{\ast} \leq K^{\ast}\) and in this case \([K:L] =[K^{\ast}:L^{\ast}]\).
\(L \triangleleft K \iff L^{\ast} \triangleleft K^{\ast}\), and in this case, \(K / L \cong K^{\ast} / L^{\ast}\).

Proof: Let \(A = \{ L: H\leq L \leq G \}\) and \(B = \{ \tilde{G}: \tilde{G} \leq G / H \}\). Define \(\Phi:A \to B\) such that \(\Phi(L) = L / H\). Suppose that \(L,K \in A\) and \(L / H = K /H\). We shall prove that \(L = K\). Let \(\ell \in L\). Then, \(\ell H \in L / H\), i.e., there exists \(k \in K\) such that \(\ell H = k H\). Thus, \(\ell ^{-1} k \in H \subseteq K\). Since \(k \in K\) and \(\ell^{-1}k \in K\), we have \(\ell = k \cdot (\ell^{-1}k)^{-1} \in K\), whence \(L \subseteq K\). The other inclusion is the same. Hence \(\Phi\) is injective. Now, let \(C \in B\), i.e., \(C \leq G / H\). Let \(K = p ^{-1}(C) \leq G\) and \(\operatorname{Ker}(p) \subseteq K\). Then, \(H \subseteq K\). Thus, \(\Phi(K)=C\). Conclude that \(\Phi\) is a bijection. We shall now prove points 1 and 2:

It is clear that \(L \leq K \iff L ^\ast \leq K^{\ast}\). To prove that the indices coincide, it suffices to prove that there is a bijection between the cosets of \(K/L\) and \(K^\ast/L^{\ast}\). Let \(f:K/L \to K^{\ast}/L^{\ast}\) be such that \(kL \mapsto p(k)L^{\ast} = kH (L / H)\).
- Well-definedness of \(f\): If \(k_{1}, k_{2} \in K\) are such that \(k_{1}L = k_{2}L\), then \(k_{2}^{-1} k_{1} \in L\) and consequently \(k_{2}^{-1} k_{1} H \in L/H = L^{\ast}\), which implies that \(f(k_{1}L) = p(k_{1}) L^\ast = p(k_{2}) L^{\ast} = f(k_{2}L)\), and therefore \(f\) is well defined.
- Injectivity of \(f\): Let \(k_{1}, k_{2} \in K\) be such that \(f(k_{1}L) = f(k_{2}L)\). Note that \(f\) is injective, since
  \[\begin{aligned} p(k_{1})L^{\ast} = p(k_{2}) L^{\ast} &\implies p(k_{2})^{-1} \ast p(k_{1}) \in L^{\ast} \implies p(k_{2}^{-1} k_{1}) \in L^\ast = L/H \\ k_{2}^{-1} k_{1} H \in L / H &\implies k_{2}^{-1} k_{1} \in L \implies k_{1}L = k_{2}L, \end{aligned}\]
- Surjectivity of \(f\): This follows directly from the surjectivity of \(p\). Thus, \([K:L] = [K^{\ast}:L^{\ast}]\).
The isomorphism between \(K/L\) and \(K^{\ast}/L^{\ast}\) follows from the third isomorphism theorem.