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Let \(\psi: G\to G'\) be a group homomorphism with kernel \(N\). Then \(G/N\) and \(\psi(G)\) are isomorphic. The canonical isomorphism \(\mu: G/N \to \psi(G)\) is given by \(\mu(gN) = \psi(g)\).
See here.
Let \(H\) and \(N\) be two subgroups of a group \(G\). The join \(\langle H, N\rangle\) of \(H\) and \(N\) is the smallest subgroup of \(G\) that contains both \(H\) and \(N\).
Equivalently, \(\langle H, N\rangle\) is the intersection of all subgroups of \(G\) containing \(H\) and \(N\).
Let \(H\) and \(N\) be two subgroups of a group \(G\). If \(N\) is normal, then \(\langle H,N\rangle = HN = NH\). Especially, \(HN\) is a subugroup of \(G\).
Proof
\(HN\) clearly contains both \(H\) and \(N\). Thus \(\langle H, N\rangle\subset HN\).
Any subgroup containing both \(H\) and \(N\) must also contain all elements of the form \(hn, h\in H, n\in N\). In particular, we have \(HN \subset \langle H, N\rangle\). ◼
兩個 finite subgroup 的 join 可能 infinite！
因為 \(H\cup N\) 不見得是 subgroup，所以定義 join（\(H\cap N\) 是 subgroup）。
另外，join 也可以如下表示：
\[\langle H,N\rangle = HN \cup HNHN \cup \cdots = \bigcup^{\infty}_{i=1}(HN)^i。\]Let \(H\) be a subgroup of \(G\) and \(N\) be a normal subgroup of \(G\). Then \((HN)/N \cong H/(H\cap N)\).
Proof
這種有 coset 又有 isomorphism 的，通常利用 the first isomorphism theorem。
Define a function \(\phi: H\to (HN)/N\) by \(\phi(h)=hN\). We claim that,
\(\phi\) is a homomorphism:
This is obvious since \(N\) is normal.
\(\text{ker}(\phi) = H\cap N\):
If \(h\in \text{ker}(\phi)\), then \(hN = N\) and thus \(h\in N\). Since \(h\) is in \(H\) by definition, we have \(h\in H\cap N\). Conversely, every element in \(H\cap N\) is also in \(\text{ker}(\phi)\). Thus \(\text{ker}(\phi) = H\cap N\).
\(\phi\) is onto:
Given \(hnN\in HN/N\) with \(h\in H\) and \(n\in N\). Then \(\phi(h) = hN = hnN\). Thus \(\phi\) is onto.
By the first isomorphism theorem, \((HN)/N \cong H/(H\cap N)\). ◼
我們也可以構造 \(\rho: HN\to H/(H\cap N)\) 來證明，但上述的 \(\phi\) 比較方便（因為 \(H\) 比較小）。
另一重點是，我們不需檢查 \(\phi\) 是否 well-defined，因為 \(\phi\) 定義在 \(H\) 上，而非 quotient group 上。
從上述定理，我們可以得知當 \(\vert G\vert < \infty\)，
\[\vert HN\vert = {\vert N\vert \vert H\vert \over \vert H\cap N\vert}。\]若 \(\vert H\cap N\vert =1\)，則 \(\vert HN\vert =\vert N\vert \vert H\vert\)，亦即所有 \(HN\) 內的元素可以唯一被 \(x\in H\) 和 \(y\in N\) 的乘積表示。定理如下：
Let \(G\) be a finite group with a subgroup \(H\) and a normal subgroup \(N\). If \(\vert H\cap N\vert = 1\) and \(\vert G\vert = \vert N\vert \vert H\vert\), then every element of \(G\) can be written as \(xy\) for some unique \(x\in H\) and \(y\in N\).
If \(\gcd(\vert H\vert ,\vert N\vert )=1\), then we always have \(\vert H\cap N\vert = 1\).
Example
Let \(G\) be a group of order \(2p\) where \(p\) is an odd prime. By Cauchy’s Theorem, there exists an element \(a\) of order \(2\) and an element \(b\) of order \(p\). Since \(\langle b\rangle\) is of index \(2\), it is a normal subgroup. In this case, we have
\[\langle a,b\rangle = \langle a\rangle\langle b\rangle = G.\]In other words,
\[G = \{e, b, b^2, \cdots, b^{p-1}, a, ab, ab^2, \cdots, ab^{p-1}\}.\]Let \(H\) and \(K\) be normal subgroups of \(G\) with \(K\leq H\). Then \(G/H \cong (G/K)/(H/K)\).
Proof
Define \(\phi:G/K \to G/H\) by \(\phi(gK) = gH\). We claim that
Then by the first isomorphism theorem, we have proved this theorem. ◼
Q: Let \(G = \Bbb Z^2\) and \(H=\langle(1, 2), (2, 1)\rangle\). Find the structure of \(G/H\).
Solution \(\rm I\)
Let \(K=\langle(3, 0),(0, 3)\rangle\), which is a subgroup of \(H\). It is easy to see that \(G/K \cong \Bbb Z_3 \times \Bbb Z_3 = \{K, [(0,1)+K], [(0,2)+K],\cdots,[(2,2)+K]\}\). Moreover, we can see that
\[H/K = \{K, [(1,2)+K],[(2,1)+K] \} \cong \Bbb Z_3.\]By the above theorem, \(\vert G/H\vert = {\vert G/K\vert\over\vert H/K\vert} = 3\), which implies that \(G/H\cong \Bbb Z_3\).
Solution \(\rm II\)
change of basis ?