2.4 Fundamental Isomorphism
Theorem 2.4.1 (\(1^{\text {st}}\) Isomorphism Theorem). Let \(G\) and \(H\) be groups, and let \(Q:\, G\longrightarrow H\) be a homomorphism
from \(G\) to \(H\), where \(ker Q = K\), Then \(\, G\big /K\, \cong \, H\).
Proof. Define the map \(\phi :\, G\big /K \, \longrightarrow H\) by \(\phi (Ka) = Q(a)\) for each \(Ka\in G\big /K\). We first show that \(\phi \) is well defined. If \(Ka_1 = Ka_2\), then \(Ka_1 = a_2\) for some \(k\in K = ker Q\).
So, \(\, Q(Ka_1) = Q(a_2)\). But
\[Q(Ka_1) = Q(K)\, Q(a_1) = e\, Q(a_1) = Q(a_1)\]
So that \(\, Q(a_1) = Q(a_2)\). Therefore, \(Q(a)\) is determined solely by the cosets of \(K\) to which \(a\) belongs. So \(\phi \) is well
defined.
Next, we show that \(\phi \) preserves the operation (it is a homomorphism). Let \(Ka\, ,\, Kb\in G\big /K\, , \) then \begin {align*} \phi \big ((Ka)(Kb)\big ) & = \phi \big (K(ab)\big )\\ & = Q(ab) = Q(a)\, Q(b)\\ & = \phi (Ka)\, Q(Kb)\hspace {0.5cm}\text {as required} \end {align*}
Clearly, \(\phi \) is onto since \(Q\) is. Finally, we show that \(\phi \) is one-to-one. i.e , \(ker\, \phi =\) identity \(= \{Ke\}\). Indeed \(ker\, \phi = \{Ke\}\) because, if \(Ka \in ker\, \phi \), then
\(Q(a) = \phi (Ka) = e\) and therefore, \(a\in ker\, Q = K\) i.e \(Ka = K = Ke\). Thus \(\phi \) is one-to-one.
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- 1.
- Every Cyclic group is a homomorphic image of \(\mathbb {Z}\). i.e, Thus \(G\cong \mathbb {Z}\) or \(\, G\cong \mathbb {Z}\big /n\mathbb {Z}\,\) for some \(n\).
Because, we may write \(\, G = \langle g\rangle \,,\) if \(\begin {vmatrix} g\\ \end {vmatrix} = \infty \,, \) then \(\, G\cong \mathbb {Z}\); if \(\begin {vmatrix} g\\ \end {vmatrix} = n\), then \(\, G\cong \mathbb {Z}\big /n\mathbb {Z}\). - 2.
- Consider the epimorphism det\(:\, G L_n(\mathbb {R})\longrightarrow \mathbb {R}^x\). Then \(ker(det) = SL_n(\mathbb {R})\). Then
\[GL_n(\mathbb {R})\big /SL_n(\mathbb {R}) \, \cong \, \mathbb {R}^x\]
Consider the group \(G = \Big \{\begin {pmatrix} a & b\\ 0& c\\ \end {pmatrix}\, , \, a\, , \, b\, , \, c\in \mathbb {R}\, , \, ac\neq 0\Big \}\) and \(b\in \mathbb {R}^x\) (\(b\) is non-negative).
Now, consider a map \(\, f:\, G\longrightarrow \mathbb {R}^x\), defined by \(f\Big (\begin {pmatrix} a & b\\ 0 & c\\ \end {pmatrix}\Big ) = ac\).- \(-\)
- \(f\) as defined is a hormomorphism since, for \(\begin {pmatrix} a & b\\ 0 & c\\ \end {pmatrix}\, , \, \begin {pmatrix} x & y\\ 0 & z\\ \end {pmatrix}\in G\),
\[f\begin {pmatrix} \begin {pmatrix} a & b\\ 0 & c\\ \end {pmatrix} \begin {pmatrix} x & y \\ 0 & z\\ \end {pmatrix} \end {pmatrix} = \begin {pmatrix} ax & ay + bz\\ 0 & cz\\ \end {pmatrix} = ax \, cz = ac\, xz = f\begin {pmatrix} a & b \\ 0 & c\\ \end {pmatrix}\cdot f\begin {pmatrix} x & y\\ 0 & z\\ \end {pmatrix}\]
- \(-\)
- \(f\) is onto (epimorphism).
for each \(y\in \mathbb {R}^x\, \exists \, \) a matrix \(\, \begin {pmatrix} y & *\\ 0 & 1\\ \end {pmatrix}\, \ni \, f\begin {pmatrix} \begin {pmatrix} y & *\\ 0 & 1\\ \end {pmatrix} \end {pmatrix} = y\cdot 1 = y\).
- \(-\)
- \(f\) is one-to-one i.e \(\, kern(f) = \Big \{\begin {pmatrix} a & b\\ 0 & a^{-1}\\ \end {pmatrix}\Big |\, a\, , \, b\in \mathbb {R}\Big \} = k\)
Clearly, \(\,\, G\big /K\,\, \cong \, \, \mathbb {R}^x\) \begin {align*} GK & = f\begin {pmatrix} \begin {pmatrix} a & b\\ 0 & c\\ \end {pmatrix} \begin {pmatrix} x & y\\ 0 & x^{-1}\\ \end {pmatrix} \end {pmatrix} = f\begin {pmatrix} ax & ay + bx^{-1}\\ 0 & cx^{-1}\\ \end {pmatrix}\\\\ & = acxx^{-1} = ac\in \mathbb {R}^x\\ \end {align*}
- 3.
- Consider a symmetric group \(S_n\) and the homomorphism \(Sg_n :\, S_n \longrightarrow \{-1\,, \, 1\}\), has its kernel as \(A_n\). Thus \(\,S_n\big /A_n \, \cong \, \{-1\, , \, 1\}\)
Theorem 2.4.3 (\(2^{\text {nd}}\) Isomorphism Theorem). Let \(G\) be a group. Let \(H\leq G\) and \(K\trianglerighteq G\). Then \(\, \, H K\big / K\, \cong \, H\big / H\cap K\).
Proof. Routine verification shows that \(HK\) is a group and that \(K\trianglerighteq HK\) and that \(H\cap K\trianglerighteq H\). Now, consider the map
\(H\longrightarrow HK\big / K\hspace {0.3cm} h\longrightarrow hk\). This map has ker\(H\cap K\) is surjuctive. So b the first isomorphism theorem, \(\, \, H\big /H\cap K\, \, \cong \, \, HK\big /K\).
Theorem (\(3^{\text {rd}}\) Isomorphism Theorem)“Absorption property of Cosets”.
Let \(G\) be a group. Let \(K \trianglelefteq G\), and let \(N\leq K\). Then \(\, K\big /N\trianglelefteq G\big /N\,\) and \(\, G/N\Big /K/N\, \cong \, G\big /K\)
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Proof. Consider the map \(\, G\big /N\longrightarrow G\big /K\,\) defined by \(gN\longmapsto g K\). This is well defined since, if \(\, g'N = g N\,\) then \(g' = gn\) for some \(n\in \textbf {N}\). Now, since
\(N\subset K\) it implies that \(g'K = gk\). The map is indeed a homomorphism because \(\, gN\, g'N = g g'N \longmapsto g\, g' K = gk\, g'k\). The map is clearly surjective.
The map has the Kernel \(K/N\). Thus by the first isomorphism theorem, \(\, \big (G/N\big )\Big /\big (K/N\big )\, \cong \, G\big / K\)
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