9.1: The First Isomorphism Theorem
A very powerful theorem, called the First Isomorphism Theorem , lets us in many cases identify factor groups (up to isomorphism) in a very slick way. Kernels will play an extremely important role in this. We therefore first provide some theorems relating to kernels.
Theorem \(\PageIndex{1}\)
Let \(G\) and \(G'\) be groups, let \(\phi\) be a homomorphism from \(G\) to \(G'\text{,}\) and let \(K=\text{Ker}\; \phi\text{.}\) Then for \(a,b\in G\text{,}\) \(aK=bK\) if and only if \(\phi(a)=\phi(b)\text{.}\)
- Proof
-
Let \(a,b∈G\). Then
\( \begin{array} &ϕ(a)=ϕ(b)&⇔ϕ(b)^{−1}ϕ(a)=e_{G′}\\&⇔ϕ(b^{−1}a)=e_{G′}\\&⇔b^{−1}a∈K\\&⇔a∈bK\\&⇔aK=bK, \end{array}\)
as desired.
Corollary \(\PageIndex{1}\)
Let \(\phi\) be a homomorphism from group \(G\) to group \(G'\text{.}\) Then \(\phi\) is one-to-one (hence a monomorphism) if and only if \(\text{Ker} \phi=\{e_G\}\text{.}\)
- Proof
-
Clearly, if \(ϕ\) is one-to-one then \(\text{Ker}ϕ=\{e_G\}\). Conversely, assume \(\text{Ker}ϕ=\{e_G\}\).
If \(a,b∈G\) with \(ϕ(a)=ϕ(b)\), then by the above theorem, \(a\text{Ker}ϕ=b\text{Ker}ϕ\). But \(a\text{Ker}ϕ=a\{e_G\}=\{a\}\) and \(b\text{Ker}ϕ=b\{e_G\}=\{b\}\). Thus, \(a=b\), and we see that \(ϕ\) is one-to-one.
We now prove a theorem that provides the meat and potatoes of the First Isomorphism Theorem.
Theorem \(\PageIndex{2}\): Factorization Theorem
Let \(G\) and \(G'\) be groups, let \(\phi\) be a homomorphism from \(G\) to \(G'\text{,}\) let \(K=\text{Ker} \phi\text{,}\) let \(N\) be a normal subgroup of \(G\) with \(N\subseteq K\text{,}\) and let \(\Psi\) be the canonical epimorphism from \(G\) to \(G/N\text{.}\) Then the map \(\overline{\phi}: G/N \to G'\) defined by \(\overline{\phi}(aN)=\phi(a)\) is a well-defined homomorphism, with \(\overline{\phi} \circ \Psi=\phi\text{.}\)
We summarize this in the following picture:
- Proof
-
We define \(\overline{\phi}:G/N→G′\), as indicated above, by \(\overline{\phi}(aN)=ϕ(a)\) for every \(aN∈G/N\). Since \(\overline{\phi}\) is defined using coset representatives, we must first show that \(\overline{\phi}\) is well-defined. So let \(aN=bN∈G/N\). Then \(\overline{\phi}(aN)=ϕ(a)\) and \(\overline{\phi}(bN)=ϕ(b)\), so we must show that \(ϕ(a)=ϕ(b)\). Since \(aN=bN\) and \(N⊆K\), we have that \(aK=bK\) by Statement 6 of Theorem \(7.2.3\); thus \(ϕ(a)=ϕ(b)\) (using Theorem \(9.1.1\)). So \(\overline{\phi}\) is well-defined.
Next, we show that \(\overline{\phi}\) is a homomorphism. Let a\(aN,bN∈G/N\). Then
\(\overline{\phi}((aN)(bN))= \overline{\phi}(abN)=ϕ(ab)=ϕ(a)ϕ(b)=\overline{\phi}(aN)\overline{\phi}(bN)\).
Finally, for every \(a∈G\),
\((\overline{\phi} \circ Ψ)(a)=\overline{\phi}(Ψ(a))=\overline{\phi}(aN)=ϕ(a)\);
so \(\overline{\phi} \circ Ψ=ϕ\) , as desired.
Note that the above theorem does not state that \(\overline{\phi}\) is a monomorphism or an epimorphism. This is because in general it may be neither! We do have the following theorem:
Theorem \(\PageIndex{3}\)
Let \(G\text{,}\) \(G'\text{,}\) \(\phi\text{,}\) \(N\text{,}\) and \(\overline{\phi}\) be as defined in the Factorization Theorem. Then
- \(\overline{\phi}\) is onto (an epimorphism) if and only if \(\phi\) is onto (an epimorphism); and
- \(\overline{\phi}\) is one-to-one (a monomorphism) if and only if \(N=\text{Ker} \phi\text{.}\)
- Proof
-
This is clear, since \(\overline{\phi}(G/N)=ϕ(G)\).
By Corollary \(9.1.1\), \(\overline{\phi}\) is one-to-one if and only if \(\text{Ker}\overline{\phi}=\{N\}\). But
\(\begin{array}& \text{Ker}\overline{\phi}&=\{aN∈G/N:\overline{\phi}(aN)=e_{G′}\} \\&=\{aN∈G/N:ϕ(a)=e_{G′}\} \\&=\{aN∈G/N:a∈\text{Ker}ϕ\}. \end{array}\)
So \(\text{Ker}\overline{\phi}=\{N\}\) if and only if
\(\{aN∈G/N:a∈\text{Ker}ϕ\}=\{N\}\),
in other words if and only if \(aN=N\) for all \(a∈\text{Ker}ϕ\). But
\(\begin{array} &&3&&aN=N &\text{ for all } a∈\text{Ker}ϕ \\&& ⇔&a∈N &\text{ for all } a∈\text{Ker}ϕ \\&&⇔&\text{Ker}ϕ⊆N \\&&⇔&\text{Ker}ϕ=N, \end{array}\)
since we are given that \(N⊆\text{Ker}ϕ\). Thus, \(\overline{\phi}\) is one-to-one if and only if \(N=\text{Ker}ϕ\), as desired.
We are now ready to state the all-important First Isomorphism Theorem, which follows directly from the Factorization Theorem and Theorem \(9.1.3\).
Theorem \(\PageIndex{4}\): First Isomorphism Theorem
Let \(G\) and \(G'\) be groups, with homomorphism \(\phi:G \rightarrow G'\text{.}\) Let \(K=\text{Ker} \phi\text{.}\) Then \(G/K \simeq \phi(G)\text{.}\) In particular, if \(\phi\) is onto, then \(G/K\simeq G'\text{.}\)
So to prove that a factor group \(G/N\) is isomorphic to a group \(G'\text{,}\) it suffices to show there exists an epimorphism from \(G\) to \(G'\) that has \(N\) as its kernel.
Example \(\PageIndex{1}\)
Letting \(n\in \mathbb{Z}^+\text{,}\) let's identify a familiar group to which \(GL(n,\mathbb{R})/SL(n,\mathbb{R})\) is isomorphic. As in Example \(8.2.3\), the map \(\phi:GL(n,\mathbb{R})\to \mathbb{R}^*\) defined by \(\phi(A)\) is a homomorphism with kernel \(SL(n,\mathbb{R})\text{.}\) Moreover, \(Phi\) clearly maps onto \(\mathbb{R}^*\text{:}\) indeed, given \(\lambda \in \mathbb{R}^*\text{,}\) the diagonal matrix having \(\lambda\) in the uppermost left position and \(1\)'s elsewhere down the diagonal gets sent to \(\lambda\) by \(\phi\text{.}\) So by the First Isomorphism Theorem, we have \(GL(n,\mathbb{R})/SL(n,\mathbb{R}) \simeq \mathbb{R}^*\text{.}\)
Example \(\PageIndex{2}\)
Let \(G=S_3\times \mathbb{Z}_{52}\) and let \(N=S_3 \times \{0\}\subseteq G\text{.}\) It is straightforward to show that \(N\) is normal in \(G\text{.}\) What is the structure of \(G/N\text{?}\) Well, define \(\phi:G\to \mathbb{Z}_{52}\) by \(\phi((\sigma, a))=a\text{.}\) Then \(\phi\) is clearly an epimorphism and \(\text{Ker} \phi=\{(\sigma,a)\in G\,:a=0\}=N\text{.}\) So \(G/N\) is isomorphic to \(\mathbb{Z}_{52}\text{.}\)
Generalizing the above example, we have the following theorem, whose proof we leave to the reader.
Theorem \(\PageIndex{5}\)
Let \(G=G_1\times G_2 \times \cdots \times G_k\) (where \(k\in \mathbb{Z}^+\)) and let \(N_i\) be a normal subgroup of \(G_i\) for each \(i=1,2,\ldots, k\text{.}\) Then \(N=N_1 \times N_2 \times \cdots \times N_k\) is a normal subgroup of \(G\text{,}\) with \(G/N \simeq G_1/N_1 \times G_2/N_2 \cdots \times G_k/N_k.\)
We provide one more cool example of using the First Isomorphism Theorem. Clearly, since \(\mathbb{R}\) is abelian, \(\mathbb{Z}\) is a normal subgroup of \(\mathbb{R}\text{.}\) What is the structure of \(\mathbb{R}/\mathbb{Z}\text{?}\) Well, in modding \(\mathbb{R}\) out by \(\mathbb{Z}\) we have essentially identified together all real numbers that are an integer distance apart. So we can think of the canonical epimorphism from \(\mathbb{R}\) to \(\mathbb{R}/\mathbb{Z}\) as wrapping up \(\mathbb{R}\) like a garden hose! Thus, one might guess that \(\mathbb{R}/\mathbb{Z}\) has some circle-like structure—but if we want to think of it as a group, we have to figure out what the group structure on such a “circle” would be!
We leave, for a moment, our group \(\mathbb{R}/\mathbb{Z}\text{,}\) and look at how we can consider a circle to be a group.
Recall that for every \(\theta \in \mathbb{R}\text{,}\) \(e^{i\theta}\) is defined to be \(\cos \theta + i\sin \theta\text{.}\) It is clear then that the set \(S^1=\{e^{i\theta} \,:\, \theta\in \mathbb{R}\}\) is the unit circle in the complex plane.
Remark
Note that if \(\theta_1, \theta_2\in \mathbb{R}\text{,}\) then \(e^{i\theta_1}=e^{i\theta_2}\) if and only if \(\theta_1-\theta_2 \in 2\pi \mathbb{Z}\text{.}\)
Theorem \(\PageIndex{6}\)
\(S^1\) is a group under the multiplication \(e^{i\theta_1}e^{i\theta_2}=e^{i(\theta_1+\theta_2)}\text{.}\)
- Proof
-
The tricky part is showing that the operation is well-defined. Suppose \(θ_1,θ_2,t_1\), and \(t_2\) are in \(R\), with \(e^{iθ_1}=e^{it_1}\) and \(e^{iθ_2}=e^{it_2}\). We want to show that
\(e^{iθ_1}e^{iθ_2}=e^{it_1}e^{it_2}\),
i.e., that
\(e^{i(θ_1+θ_2)}=e^{i(t_1+t_2)}\).
Now, \(e^{iθ_1}=e^{it_1}\) and \(e^{iθ_2}=e^{it_2}\) imply that \(θ_1−t_1=2πm\) and \(θ_2−t_2=2πn\) for some \(m\) a nd \(n\) in \(\mathbb{Z}\); hence,
\(θ_1+θ_2−(t_1+t_2)=2π(m+n)∈2π\mathbb{Z}\).
Thus, \(e^{i(θ_1+θ_2)}=e^{i(t_1+t_2)}\), so our operation is well-defined.
Since products of elements in \(S^1\) are in \(S^1\), we have that \(S^1\) is a binary structure; moreover, associativity of the operation in \(S^1\) follows from the associativity of addition in \(\mathbb{R}\). Moreover, \(e^{i0}=1\) clearly acts as an identity element in \(S^1\), and if \(e^{iθ}∈S^1\), then \(e^{iθ}\) has inverse \(e^{i(−θ)}∈S^1\). So \(S^1\) is a group under the described operation.
Beautifully, it turns out that our group \(\mathbb{R}/\mathbb{Z}\) is isomorphic to \(S^1\text{.}\)
Theorem \(\PageIndex{7}\)
\(\mathbb{R}/\mathbb{Z} \simeq S^1\text{.}\)
- Proof
-
We will define an epimorphism \(ϕ\) from \(\mathbb{R}\) to \(S^1\) with \(\text{Ker}ϕ=\mathbb{Z}\); then we'll have \(\mathbb{R}/\mathbb{Z}≃S^1\), by the First Isomorphism Theorem.
Define \(ϕ:\mathbb{R}→S^1\) by \(ϕ(r)=e^{i2πr}\). We have that \(ϕ\) is a homomorphism, since for every \(r,s∈\mathbb{R}\), we have
\(ϕ(r+s)=e^{i2π(r+s)}=e^{i2πr+i2πs}=e^{i2πr}e^{i2πs}=ϕ(r)ϕ(s)\).
Moreover, \(ϕ\) is clearly onto, since if \(e^{iθ}∈S^1\), then
\(e^{iθ}=e^{i2π(θ2π)}=ϕ(θ2π)\).
Finally, \(\text{Ker}ϕ=\mathbb{Z}\): indeed,
\(\begin{array}& r∈\text{Ker}ϕ&⇔ϕ(r)=1\\&⇔e^{i2πr}=1\\&⇔\cos2πr+i\sin2πr=1\\&⇔\cos2πr=1 \text{ and } \sin2πr=0\\&⇔r∈\mathbb{Z}. \end{array}\)
Thus, \(\text{Ker}ϕ=\mathbb{Z}\), and hence \(\mathbb{R}/\mathbb{Z}≃S^1\), as desired.