群的单同态的推出(pushout)

以下定理在我学习低维拓扑时遇到(推出的构造来自Van Kampen定理), 但并没找到详细证明. 我在这里给出一个不那么繁琐的证明过程.

Theorem 1

If \(A\to B, A\to C\) are both group monomorphisms, then the pushout \(B\to B\sqcup_{A} C, C\to B\sqcup_{A} C\) are also monomorphisms.

(假装这里有一个交换图表)

Proof

We attempt to construct a group satisfying the properties of pushout.

Step 1 Define the group as a set.

(by Choice Axiom) \(C = \bigsqcup_{c_\alpha\in \Lambda_C} c_\alpha f(A), B = \bigsqcup_{b_\alpha\in \Lambda_B} b_\alpha g(A)\) are divided into some cosets. WLOG \(1_C\in \Lambda_C, 1_B\in \Lambda_C\), Let \(\Lambda = \Lambda_B\sqcup \Lambda_C\), and define a boolean map \(\mathbf{\delta}:\Lambda\times\Lambda\to\{0,1\}\), such that \(\delta(g,h)=1\) if and only if \(g\) and \(h\) lie in \(\Lambda_B, \Lambda_C\), respectively.

Let \(G\) consists of all words of the form \(g_1\cdots g_na\) such that

(C1) \(n\ge 0, g_i\in \Lambda-\{1_B,1_C\}, a\in A\),

(C2) \(\delta(g_i,g_{i+1})=1\).

Step 2 Define the multiplication.

Consider

\[g_1\cdots g_{n_1}a \times h_1\cdots h_{n_2}b, \]

where \(g_i, h_i\in \Lambda-\{1_B,1_C\}, a,b\in A\). We'll define the product inductively(on \(n_2\)).

Case 2.1
For \(x\in A\), define

\[g_1\cdots g_{n_1}a\times x = g_1\cdots g_{n_1}(ax). \]

Case 2.2
For \(x\in \Lambda - \{1_B, 1_C\}\),

Case 2.2.1
If \(n_1=0\) or \(n_1>0, \delta(g_{n_1}, x)=1\), then WLOG \(x\in \Lambda_B\), calculate \(f(a)x = x'f(a')\) (obviously \(x'\neq 1_B\)) and define

\[g_1\cdots g_{n_1}a\times x = g_1\cdots g_{n_1}x'a' \]

Case 2.2.2
If \(n_1>0, \delta(g_{n_1}, h_1)=0\), WOLG \(g_{n_1}, h_1 \in \Lambda_B\), calculate \(g_{n_1}f(a)x = x'f(a')\). Let \(X = \begin{cases} a', &x'=1_B;\\ x'a', &x'\neq 1_B. \end{cases}\) and define

\[g_1\cdots g_{n_1}a\times x = g_1\cdots g_{n_1-1}X \]

Now define the following word as the product by induction
\begin{equation}
g_1\cdots g_{n_1}a \times h_1\cdots h_{n_2}b =
\begin{cases}
(g_1\cdots g_{n_1}a \times h_1) \times h_2\cdots h_{n_2}b, &n_2>0;\
g_1\cdots g_{n_1}a \times b, &n_2=0.
\end{cases}
\end{equation}

Step3 Verify the associative law.

Consider

\[(f_1\cdots f_{n_1}a \times g_1\cdots g_{n_2}b) \times h_1\cdots h_{n_3}c, \]

\[f_1\cdots f_{n_1}a \times (g_1\cdots g_{n_2}b \times h_1\cdots h_{n_3}c). \]

We'll also show that they are equal by induction on \((n_3, n_2)\).
Notice that for \(g_i\in \Lambda_B\sqcup A\), \(g_1\times\cdots\times g_m\) does not depend on the order of the multiplication since the product in \(B\) satisfies the associative law. For example, we have

\[(a_1a_2)\times g_2 = a_1\times (a_2\times g_2), g_1(a_1a_2)\times g_2 = g_1a_1\times (a_2\times g_2). \]

that is to say, for any \(F\in G, a\in A, g\in \Lambda\),

\[(F\times a)\times g = F\times (a\times g). \]

Denote \(f_1\cdots f_{n_1}a\) as \(F\).

Case 3.1
If \(n_2=n_3=0\), then

\[(f_1\cdots f_{n_1}a \times b) \times c = f_1\cdots f_{n_1}(abc) = f_1\cdots f_{n_1} a\times (b\times c). \]

Case 3.2
If \(n_2>0, n_3=0\), then by induction hypothesis on \((0, n_2-1)\),

\[\begin{aligned} &(F \times g_1\cdots g_{n_2}b) \times c = ((F \times g_1) \times g_2\cdots g_{n_2}b) \times c \\= &(F \times g_1) \times (g_2\cdots g_{n_2}(bc)) = F \times (g_1\cdots g_{n_2}b \times c). \end{aligned} \]

Case 3.3
If \(n_3>0, n_2=0\), by induction hypothesis on \((n_3-1, 1)\), we have

\[\begin{aligned} &(F \times b) \times h_1\cdots h_{n_3}c = ((F \times b) \times h_1) \times h_2\cdots h_{n_3}c \\= &(F \times (b \times h_1)) \times h_2\cdots h_{n_3}c \\= &F \times ((b \times h_1) h_2\cdots h_{n_3}c)= F \times (b \times h_1\cdots h_{n_3}c). \end{aligned} \]

Case 3.4
If \(n_3>0, n_2>0\), then by induction hypothesis on \((n_3, n_2-1)\), \((n_3-1, n_2')\) and \((n_3-1, n_2'+1)\),

\[\begin{aligned} &(F \times g_1\cdots g_{n_2}b) \times h_1\cdots h_{n_3}c = ((F \times g_1) \times g_2\cdots g_{n_2}b) \times h_1\cdots h_{n_3}c \\= &(F \times g_1)\times (g_2\cdots g_{n_2}b \times h_1\cdots h_{n_3}c) = (F \times g_1)\times ((g_2\cdots g_{n_2}b\times h_1) \times h_2\cdots h_{n_3}c) \\= &((F \times g_1)\times (g_2\cdots g_{n_2}b\times h_1)) \times h_2\cdots h_{n_3}c = (F \times (g_1\cdots g_{n_2}b\times h_1)) \times h_2\cdots h_{n_3}c \\= &F\times ((g_1\cdots g_{n_2}b \times h_1) \times h_2\cdots h_{n_3}c) = F\times (g_1\cdots g_{n_2}b \times h_1\cdots h_{n_3}c). \end{aligned} \]

Now \(G\) is actually a group since every element in \(\Lambda\cup A\) has inverse element in \(G\).

Step 4 Verify the properties of pushout.

Since the elements in \(G\) is made up of elements in \(A, B\) and \(C\), there're natural injections from \(A,B,C\) to \(G\). There's also a natural homomorphism from \(G\to B\ast C/N, x\mapsto\begin{cases} f(x)+N = g(x)+N, &x\in A;\\ x+N, &x\in \Lambda;\\ \end{cases}\). Thus, \(G\) inherits the property of pushout.

Now the injectivity of \(B\to B\sqcup_{A} C, C\to B\sqcup_{A} C\) is obvious.

\[\newcommand{\laa}{\langle\hspace{-0.2em}\langle} \newcommand{\raa}{\rangle\hspace{-0.2em}\rangle} \newcommand{\genr}[1]{\langle#1\rangle} \]

事实上, 这个结果不能推广到只有一边是单射的情形.

Example 2

In this example, we attempt to show that in the group category, if \(D\) is the pushout of \(f:A\to B, g:A\to C\), then \(g\) is injective cannot guarantee that \(B\to D\) is injective.

Let \(A=\Z/mn\Z\), \(B=\Z/n\Z, g:x\mapsto mx\) and denote \(c=f(1)\). Then the pushout \(D\) is just isomorphic to \(C/\laa c^m \raa\) where \(\laa c^m \raa\) is the minimum normal subgroup of \(C\) containing \(c^n\). Notice that if we let \(C\) be a simple group, then the pushout must be a trivial group and the homomorphism \(h\) must not be injective.

In order to guarantee the injectivity of \(f\), we should find a simple group containing a cyclic subgroup of a composite order. In fact, \(A_{8}\) has such a subgroup \(\langle\sigma^2\rangle\cong \Z/4\Z\), where \(\sigma\) is an order \(8\) element in \(S_{8}\).

Combining all the discussion above, we give such an example. \(A=\Z/4\Z=\genr{\alpha}, B=\Z/2\Z=\genr{\beta}, C=A_{8}, \sigma = (1\hspace{0.5em}2\cdots 8)\) with \(f(\alpha)=\beta, g(\alpha)=\sigma^2\), then \(f\) is injective while \(D=1\), which implies that \(h\) is not a monomorplism.