Table of Content
Let \(\phi\) be a homomorphism of a group \(G\) into a group \(G'\).
\(\phi(H)\) is the image of \(H\).
\(\phi^{-1}(H)\) is the preimage of \(H'\).
請嘗試證明，尤其是 subgroup 的部分。
Let \(\phi: G\to G'\) be a group homomorphism. If \(\phi\) is one-to-one then \(G\) and \(\phi(G)\) are isomorphic.
顯然 \(\phi: G \to \phi(G)\) 已經 onto。
Homoporphism \(\rho\) is one-to-one if \(\ker{\rho} = \{e\}\).
Homomorphism \(\rho\) is onto if \(\rho(G)\) contains a set of generators of \(G'\).
one-to-one 的 corollary 比較好用。
Let \(\phi: G\to G'\) be a group homomorphism.
For \(a \in G\), suppose its order is \(n\). Since \(\phi\) is homomorphism, we have
\[\phi(a^n) = \phi(e) = e' = \phi(a)^n,\]and thus
\[\text{ord}(\phi(a))\big\vert n.\]If \(\gcd(\vert G \vert, \vert G' \vert) = 1\), for all \(g\) in \(G\), we let \(n = \text{ord}(\phi(g))\) and \(m = \text{ord}(g)\).
By Lagrange’s Theorem, we have
\[n \big\vert \vert G' \vert,\ m \big\vert \vert G \vert.\]Moreover, by the definition of order, we have
\[\phi(g^m) = e',\]which implies that \(\phi(g)^m\) is identity and the order of \(\phi(g)\) divides \(m\). From this result, we can further observe that
\[n \big\vert \vert G \vert, \\ \Rightarrow n \big\vert \gcd(\vert G \vert, \vert G' \vert) = 1,\]which means that \(n\) is one. Thus for all \(\phi(g)\) in \(G'\), the order of them are one, i.e. \(\phi\) is a trivial homomorphism.
Describe all the homomorphism from \(\mathbb{Z}_6\) to \(\mathbb{Z}_8\).
Solution
Let \(\phi\) be a homomorphism from \(\mathbb{Z}_6\) to \(\mathbb{Z}_8\). Suppose \(\phi(\bar 1) = k\) for some \(k \in \mathbb{Z}_8\). Then since \(\phi\) is a homomorphism, we have, for all \(\bar m \in \mathbb{Z}_6\), \(\phi(\bar m) = \phi(\bar 1+\bar 1+\cdots+\bar 1) = m\phi(\bar 1) = mk\). In particular, we have \(\bar 6k = \phi(\bar 6) = \phi(\bar 0) = \bar 0\), which means that \(6k \equiv 0 \text{ mod } 8\), i.e. \(k \equiv 0 \text{ mod }4\). Thus, \(\phi(\bar 1) = \bar 0, \bar 4\), each of which uniquely determines a homomorphism.
In terms of kernel, if \(\phi(\bar 1) = \bar 0\), then \(\text{ker}(\phi) = \mathbb{Z}_6\); otherwise, \(\text{ker}(\phi) = \{\bar 0, \bar 2, \bar 4\} = \langle \bar 2 \rangle\).
To generalize, what is the number of all possible homomorphisms from \(\mathbb{Z}_n\) to \(\mathbb{Z}_m\)? From the above result, we see that
\[nk \equiv 0 \text{ mod m},\]which is to say
\[k \equiv (n/\gcd(n, m))^{-1} \text{ mod }(m/\gcd(n,m)).\]Let \((n/\gcd(n, m))^{-1} = a\). Therefore,
\[k = a, a+{m\over \gcd(n,m)}, a+{2m\over \gcd(n,m)}, \cdots, a+{(\gcd(n,m)-1)m\over \gcd(n,m)},\]which means that there are \(\gcd(n, m)\) homomorphisms.
The subgroup \(\phi^{-1}(\{e'\})\) is called the kernel of \(\phi\), denoted by \(\text{ker}(\phi)\).
也就是說，對於任意在 domain \(X\) 的 \(x\)，只要 \(\phi(x) = e'\)，那麼 \(x\) 就在 \(\text{ker}(\phi)\) 中！
Example
We know that
\[\text{sgn: }S_n \to \{\pm 1\}\]is a group homomorphism, and therefore \(A_n = \text{ker(sgn)}\) is a subgroup of \(S_n\).
\(A_n\) 是 even permutation 形成的 subgroup！
Let \(\psi: G\to G'\) be a group homomorphism, and let \(H = \text{ker}(\psi)\). Let \(a\in G\). Then the set
\[\{x \in G: \psi(x)=\psi(a) \} = \psi^{-1}(\{\psi(a) \})\]is the left coset \(aH\), and is also the right coset \(Ha\).
\(\psi^{-1}(\{\psi(a) \})\) 是 \(\psi(a)\) 的 preimage！（\(\psi(a)\) 來的地方。）
A group homomorphism \(\psi: G \to G'\) is one-to-one iff \(\text{ker}(\psi) = \{e\}\).
One-to-one 本來就會使 \(\text{ker}(\psi) = \{e\}\)；而當 \(\text{ker}(\psi) = \{e\}\) 時，對於任意 \(a \in G\)，他的 preimage 都是 \(aH = a\)，也就證明了 one-to-one。
Proof of the theorem
We first show that \(aH \subseteq \psi^{-1}(\{\psi(a)\})\): Let \(x=ah\in aH\). Since \(\psi\) is a homomorphism, we ahve \(\psi(x) = \psi(a)\psi(h) = \psi(a)e' = \psi(a)\). Therefore, \(x \in \psi^{-1}(\{\psi(a)\})\).
Then, we are to show that \(\psi^{-1}(\{\psi(a)\}) \subseteq aH\) : Suppose that \(x \in \psi^{-1}(\{\psi(a)\})\). We have \(\psi(x) = \psi(a)\), and thus \(\psi(a)^{-1}\psi(x) = e'\). Then \(\psi(a^{-1})\psi(x) = \psi(a^{-1}x) = e'\), which implies that \(a^{-1}x \in H\). We see that \(x \in aH\).
The proof for the right coset is similar. ◼
Example
Let \(\psi: \mathbb{Z} \to \mathbb{Z}_n\) defined by \(\psi(a) = \bar a\); then \(\text{ker}(\psi) = n\mathbb{Z}\). For each \(\bar b \in \mathbb{Z}_n\), its preimage \(\psi^{-1}(\{\bar b\}) = \{\cdots,b-n,b,b+n,\cdots \} = b + n\mathbb{Z}\) is indeed a coset.
A subgroup \(H\) of a group \(G\) is normal if its left cosets and right cosets coincide, that is, if
\[gH = Hg,\ \text{for all } g \in G.\]If \(H\) is a normal subgroup of \(G\), we denoted it as \(H \triangleleft G\).
看到 \(gH=Hg\)，我們可能會想到 abelian group：\(gx = xg\)；事實上，being normal 是比 being abelian 弱的性質，因為 abelian 要求一一對應，而 normal 只需要「整個集合」一樣就好了（像是 \(g_1H = Hg_2\)）。
另外，檢查一 subgroup \(H\) 是否 normal 只需看 group \(G\) 和 subgroup \(H\) 的 generator(s) 就好，因對於所有 \(g\in G\)，\(g\) 都可以用 generator(s) 來表示：如果 generator(s) 可以在 \(H\) 左右移動，則任意 \(h\) 拆成 generator(s) 之後也必可以在 \(H\) 左右移動；演示如下：
If \(G = \langle g_1, \cdots,g_n\rangle\), and let \(g = g_1^{k_1}\cdots g_n^{k_n}\). Then
\[\begin{align*} (g_1^{k_1}\cdots g_n^{k_n})h(g_1^{k_1}\cdots g_n^{k_n})^{-1} &= (g_1^{k_1}\cdots g_n^{k_n})h(g_n^{k_n})^{-1}\cdots(g_1^{k_1})^{-1} \\ &= (g_1^{k_1}\cdots g_{n-1}^{k_{n-1}})h'(g_{n-1}^{k_{n-1}})^{-1}\cdots(g_1^{k_1})^{-1} \\ &= \cdots \in H. \end{align*}\]If \(H = \langle h_1,\cdots, h_m\rangle\), and let \(h = h_1^{i_1}\cdots h_m^{i_m}\). Then
\[\begin{align*} g(h_1^{i_1}\cdots h_m^{i_m})g^{-1} = (gh_1^{i_1}g^{-1})\cdots(gh_m^{i_m}g^{-1}) \in H. \end{align*}\]非常好用！尤其在處理 permutaiton group 的時候。
另外，kernel 是 normal 的；而對任意 normal subgroup \(N\)，存在一 group homomorphism \(\rho\) 使得 \(\text{ker}\rho = N\)。
If \(\psi: G\to G'\) is a group homomorphism, then \(\text{ker}(\psi)\) is a normal subgroup.
From this theorem.
Let \(N\) be a subgroup of index two of \(G\). Then \(N\) is normal.
Proof
For \(g \in G\), if \(g \in N\), then \(gN = N = Ng\). If \(g \not \in N\), then \(G=N \bigsqcup gN = N \bigsqcup Ng\), and therefore \(gN = G\backslash N = Ng\). We now have \(gN = Ng\). ◼
The following are equivalent conditions for a subgroup \(H\) of a group \(G\) to be normal.
- \(ghg^{-1} \in H\) for all \(g\) in \(G\) and all \(h\) in \(H\),
- \(gHg^{-1} = H\) for all \(g\) in \(G\),
- \(gH = Hg\) for all \(g\) in \(G\),
- \(aHbH = abH\) for all \(a,b\) in \(G\).
Proof
\((1)\Rightarrow(2)\):
Suppose \(gHg^{-1} \subset H\) for all \(g \in G\). Replacing \(g\) with \(g^{-1}\), we have \(g^{-1}Hg \subset H\), which implies that \(H \subset gHg^{-1}\). We conclude that \(gHg^{-1} = H\).
\((4)\Rightarrow(1)\):
For \(g \in G\),
\[ghg^{-1}e \in gHg^{-1}H = gg^{-1}H = H.\]From this theorem, we see that for \(G/N\), the left coset of a normal subgroup \(N\) of \(G\), we have
\[aNbN = abN\]for all \(a,b \in G\). Then we can see that
\[aN * bN := abN\]is a well-defined binary operator on \(G/N\).
For \(c=a, d=b\), we have \(cNdN = cdN = abN = aNbN\).
\((G/N,*)\) is a group, called the quotient group of \(G\) by \(N\).
To prove \((G/N,*)\) be a group, we have to check three conditions: Associativity, Identity, Inverse.
Example \(\rm I\)
Let \(G = \mathbb{Z}_6\) and \(N = \{\bar 0, \bar 3\}\). The left cosets are \(N,\bar 1+N,\bar 2+N\). Moreover, we have \((\bar 1+N) + (\bar 1+N) = \bar 2+N\) and \((\bar 1+N) +(\bar 1+N) +(\bar 1+N) = N\). Therefore, \(G/N\) is a cyclic group of order \(3\) generated by \(\bar 1+N\).
Example \(\rm II\)
見 Remark。
對於 \(D_4\)，令 \(N = \langle \sigma^2 \rangle\)。欲檢查 \(N\) 是否 normal，我們只需看 generators \(\sigma\) 和 \(\tau\)。而
\[N\sigma = \{\sigma^3,\sigma \} = \sigma N,\\ N\tau = \{\tau\sigma^2, \tau \} = \tau N,\]所以 \(N\) is normal，且
\[D_4 = N \bigsqcup \sigma N \bigsqcup \tau N \bigsqcup \tau \sigma N.\]於是，\(D_4\) 的 quotient group is of order \(4\)，和 \(\mathbb{Z}_4\) 或 \(\mathbb{Z}_2\times \mathbb{Z}_2\) 同構。再來，因為 \(D_4/N\) 的每一元素都是 order \(2\)，所以 \(D_4/N \cong \mathbb{Z}_2\times \mathbb{Z}_2\)。
試著找出一個 group \(G'\) isomorphic to the quotient group \(G/H\)，因為 quotient group 不易處理和計算！
Some facts
Proof of 3.
If \(G = \langle a \rangle\), then every coset of \(H\) is of the form \(a^kH = (aH)^k\) for some integer \(k\). Thus, \(G/H = \langle aH \rangle\). ◼
\(H = \langle a^d \rangle\), \(d \vert n\).
要證明 quotient group 和某 group isomorphic，可以用 the first isomorphism theorem：
Construct \(\rho: G \to G'\), and then \(G/\text{ker} (\rho) \cong \rho(G)\).
Proof of 4.
Let \(\phi: G_1\times G_2 \to G_1\) be defined by \(\phi(g_1, g_2) = g_1\). Then we have \(\text{ker}(\phi) = \{e_1\}\times G_2\). We complete the proof by the first isomorphism theorem. ◼
Proof of 5.
Let \(\phi: G_1\times G_2 \to (G_1/H_1)\times (G_2/H_2)\) defined by \(\phi(g_1, g_2) = (g_1H_1, g_2H_2)\). By apply the first isomorphism theorem, we obtained the desired result. ◼
Classify the quotient group \(\mathbb{Z}_4\times \mathbb{Z}_2/\langle (2,0)\rangle\).
(Guess that the answer is \(\mathbb{Z}_2\times\mathbb{Z}_2\).)
Define a function \(\phi: \mathbb{Z}_4\times \mathbb{Z}_2 \to \mathbb{Z}_2\times \mathbb{Z}_2\) by
\[\phi(a \text{ mod } 4, b \text{ mod } 2) = (a \text{ mod } 2, b \text{ mod } 2),\]which is well-defined since \(2\vert 4\). We can also see that \(\phi\) is a homomorphism.
The kernel is \(\{(\bar 0, \bar 0),(\bar 2, \bar 0) \} = H\). Moreover \(\phi\) is onto since
\[(\bar 1, \bar 0) = \phi(\bar 1, \bar 0),\\ (\bar 0, \bar 1) = \phi(\bar 0, \bar 1).\]generator 都 map 到了！於是 onto。
Therefore, by the first isomorphism theorem we have
\[\mathbb{Z}_4\times \mathbb{Z}_2/\langle (2,0)\rangle \cong \mathbb{Z}_2\times\mathbb{Z}_2.\]先猜，然後建 isomorphism，最後用 theorem！
Let \(N\) be a normal subgroup of \(G\). Then \(\gamma:G \to G/N\) given by \(\gamma(g) = gN\) is a homomorphism with the kernel \(N\).
“with the kernel \(N\)” 指的是 \(N = \text{ker}(\gamma)\)！
Proof
Since \(N \triangleleft G\), we have \((aN)(bN) = (ab)N\), that is, \(\gamma(a)\gamma(b) = \gamma(ab)\), and thus \(\gamma\) is a homomorphism.
For the second part, we know that \(g \in \text{ker}(\gamma)\) iff \(gN=eN\). This means that \(g \in \text{ker}(\gamma)\) iff \(g \in N\). Thus \(N = \text{ker}(\gamma)\). ◼
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 \(\mu(gN) = \psi(g)\).
\(G/\text{ker}(\psi) \cong \psi(G) = \text{the image of }G = \text{Im}(G)\).
也稱作 the fundamental homomorphism theorem。
\(G/\text{ker}(\psi)\) 的 order 正是 left coset 的數量，又 isomorphic groups 的 order 必相等。因此
\[\vert G \vert = \vert \text{ker}(\psi) \vert \cdot \vert \text{Im}(G) \vert.\]Proof
For \(\mu\), we need to check four conditions:
proof of well-definedness
Suppose \(g_1N = g_2N\). We need to show that \(\mu(g_1N) = \mu(g_2N)\).
We know that \(g_1N = g_2N\) iff \(g_1 = g_2h\) for some \(h \in N\). Then
\[\mu(g_1N) = \psi(g_1) = \psi(g_2h) = \psi(g_2)\psi(h).\]Since \(N\) is the kernel of \(\psi\), we have \(\psi(h)=e'\). Therefore, \(\mu(g_1N) = \psi(g_2) = \mu(g_2N)\).
proof of one-to-one
Suppose that \(\mu(g_1N) = \mu(g_2N)\), which means \(\psi(g_1) = \psi(g_2)\). Then we have \(\psi(g_2^{-1}g_1) = e'\), which implies that \(g_2^{-1}g_1 \in \text{ker}(\psi) = N\). Hence, \(g_1N = g_2N\).
\(g_1N = g_2N\) iff \(g_2^{-1}g_1 \in N\).
proof of onto
For all \(g' \in \psi(G)\), we have some \(g \in G\) such that \(\psi(g) = g'\). Therefore, we have \(gN \in G/N\) such that \(\mu(gN) = \psi(g) = g'\).
proof of homomorphism
By the fourth statement of this theorem. ◼
Every abelian group with \(n\) generators is isomorphic to a quotient group of \(\mathbb{Z}^n\).
Proof
Let \(g_1\cdots g_n\) be a set of generators of an additive group \(G\). Then \(G = \{\sum a_ig_i \vert a_i \in \mathbb{Z} \}\). Let
\[\rho: \mathbb{Z}^n \to G\]defined by \(\rho((a_1,\cdots,a_n)) = \sum a_ig_i\), which is a surjective group homomorphism. By the first isomorphism theorem, \(G\) is isomorphic to \(\mathbb{Z}^n/\text{ker}(\rho)\). ◼
Let \(G=D_6\) and \(H_1 = \langle \sigma^2,\tau \rangle\). Find a group homomorphism from \(G\) with kernel equal to \(H_1\).
Solution
設該 homomorphism 為 \(\phi\)，利用本定理，\(G/H_1 \cong \phi(G)\)，也就是說，
\[\phi(G) \cong \{H_1, \sigma H_1\}。\]從上式可以看出，for all \(g\in \text{ker}(\phi)\), \(\phi(g) = H_1\)（因為 \(H_1\) 是 identity！）所以若定義 \(\phi(g) = gH_1\)，則
\[gH_1 = H_1 \iff g \in H_1,\]說明了 \(\text{ker}(\phi) = H_1\)！
According to the first isomorphism theorem, \(\mathbb{Z}^n/\text{ker}(\rho) \cong \rho(\mathbb{Z}^n)\). However, since \(\rho\) is surjective, we have \(\rho(\mathbb{Z}^n) = G\). Thus the result follows.