在看庄晓波老师的关于自由Abel群的视频时, 多次提到了直和与直积的概念, 每次一提到这些概念自己都需要重新翻看相关的概念, 说明还是掌握得不够, 因此谨以本文重新复习一下相关概念, 加深理解~
参考视频: P53 (52)Abel群、自由Abel群
1. 直积
设$(G, \cdot), (H, \cdot)$为群, 定义$$G \times H = \{ (g, h)|g \in G, h \in H) \},$$定义于$G \times H$之上的乘法运算为$$(g_1, h_1) \cdot (g_2, h_2) = (g_1 g_2, h_1 h_2).$$设$(G_\alpha, +), \alpha \in I$为一族Abel群, 定义它们的直积群(Direct Product)$(\prod_{\alpha \in I} G_\alpha, +)$如下:$$\prod_{\alpha \in I} G_\alpha = \{ (g_\alpha)_{\alpha \in I} | g_\alpha \in G_\alpha \}.$$$\forall (g_\alpha), (h_\alpha) \in \prod_{\alpha \in I} G_\alpha, (g_\alpha) + (h_\alpha) := (f_\alpha)$, 其中$\forall \alpha \in I$, $f_\alpha = g_\alpha + $$ h_\alpha$.
$\\$ 若$I$为有限集, i.e. 对于有限个Abel群$(G_1, +), \cdots, (G_n, +)$, $(\prod_{i = 1}^n G_i, +)$又记为$(G_1 \times \cdots \times G_n, +)$.
2. 外直和
设$\{ G_\alpha \}_{\alpha \in I}$为一族Abel群, 定义它们的外直和(Direct Sum) $(\oplus_{\alpha \in I} G_\alpha, +)$如下:$$\oplus_{\alpha \in I} G_\alpha := \{ (g_\alpha)_{\alpha \in I} | g_\alpha \in G_\alpha, \forall \alpha \in I \}$$且其每个元素的所有$g_\alpha$只有有限个非零. $\forall (g_\alpha), (h_\alpha) \in \oplus_{\alpha \in I} G_\alpha$, $(g_\alpha) + ( $$ h_\alpha) := (f_\alpha)$, 其中$\forall \alpha \in I, f_\alpha = g_\alpha + h_\alpha.$
注1: $\oplus_{\alpha \in I} G_\alpha \subset \prod_{\alpha \in I} G_\alpha$.
$\\$ 注2: 对于有限个Abel群$G_1, \cdots, G_n$, $G_1 \times \cdots \times G_n = G_1 \oplus $$ \cdots \oplus G_n$.
3. 内直和
设$G$为一个Abel群, $\{ G_\alpha \}_{\alpha \in I}$为$G$中的一族子群, 称$G$为$\{ G_\alpha \}_{\alpha \in I}$的内直和, 若$\forall g \in G$, $g$可唯一地表示为$g = \sum_\alpha g_\alpha$, 其中$g_\alpha \in G_\alpha$, $g_\alpha$中只有有限个非零, 此时记$G = \oplus_{\alpha \in I} G_\alpha$.
4. 内外直和之关系
设$G$为Abel群, $G = \oplus_{\alpha \in I} G_\alpha$为内直和分解, 则$G$显然同构于$G_\alpha, $$ \alpha \in I$的外直和:$$\forall g \in G, g = \sum_\alpha g_\alpha \mapsto (g_\alpha)_{\alpha \in I}.$$反之, 设$\{ G_\alpha \}_{\alpha \in I}$为一族Abel群, $G = \oplus_{\alpha \in I} G_\alpha$为外直和. 记$$i_\alpha: G_\alpha \to \oplus_{\alpha \in I} G_\alpha$$为典则嵌入(事实上我并没有查到典则嵌入的定义囧), $g \mapsto i_\alpha(g) = $$ (h_\alpha)$, 其中,$$h_\beta = \left\{\begin{matrix}
g,\ if \ \beta = \alpha \\
0, \ if \ \beta \ne \alpha
\end{matrix}.\right.$$则$G’_\alpha := i_\alpha(G_\alpha) \subset \oplus_{\alpha \in I} G_\alpha$, 从而显然$G$为$G’_\alpha, \alpha \in I$的內直和.
$\\$ $\because \forall g \in G( = \oplus_{\alpha \in I} G_\alpha), g = (g_\alpha)_{\alpha \in I} = \sum_\alpha i_\alpha(g_\alpha)$. 接下来定义群同态$$\pi_\alpha: \oplus_{\alpha \in I} G_\alpha \to G_\alpha, (g_\alpha) \longmapsto g_\alpha,$$则$\pi_\alpha \circ i_\alpha = id_{G_\alpha}, \pi_\alpha \circ i_\beta = 0, \forall \beta \ne \alpha$. 从而我们有$$\pi_\beta(g) = \pi_\beta(\sum_\alpha i_\alpha(g_\alpha)) = \sum_\alpha \pi_\beta \circ i_\alpha(g_\alpha) \\ = \pi_\beta \circ i_\beta(g_\beta) = g_\beta, \forall \beta \in I.$$$\therefore g = \sum_\alpha i_\alpha(\pi_\alpha(g))$. 也就是说, $g_\alpha$是由$g$与$\pi_\alpha$唯一确定的, 那么$$g = \sum_\alpha i_\alpha(g_\alpha)$$这样的分解自然也是唯一的.
$\\$ 由上述讨论亦可知, 内直和与外直和总是相生相伴的, 我们可以由内直和得到外直和, 也可由外直和得到内直和, 故经常不加区分地记内直和与外直和.
5. 直和, 直积与$Hom$之关系
不妨先定义函子$Hom$, 设$G, H$为两个Abel群, 定义$$Hom(G, H) = \{ f:G \to H | f \ is \ a \ group \ homomorphism. \}$$其上的加法运算为$$+: Hom(G, H) \times Hom(G, H) \to Hom(G, H), \\ (\varphi, \psi) \longmapsto \varphi + \psi,$$其中, $(\varphi + \psi)(g) := \varphi(g) + \psi(g)$.
$\\$ 设$(G_\alpha, \alpha \in I)$为一族Abel群, $H$为另一个Abel群, 则有$$\varphi: Hom(\oplus_{\alpha \in I} G_\alpha, H) \to \prod_{\alpha \in I} Hom(G_\alpha, H), \\ f \longmapsto \varphi(f) := (f_\alpha)_{\alpha \in I},$$其中$f_\alpha = f \circ i_\alpha$($i_\alpha : G_\alpha \to \oplus_{\alpha \in I} G_\alpha$为典则嵌入). 接下来证$\varphi$是一个群同构.
$\\$ $\varphi$显然为单射. 再证$\varphi$为满射: $\forall (f_\alpha)_{\alpha \in I} \in \prod_{\alpha \in I} Hom(G_\alpha,H)$, 定义$$f: \oplus_{\alpha \in I} G_\alpha \to H, \\ \sum_{\alpha \in I} g_\alpha \longmapsto \sum_{\alpha \in I} f_\alpha(g_\alpha).$$显然$f \in Hom(\oplus_{\alpha \in I} G_\alpha, H)$, 且$\varphi(f) = (f_\alpha)_{\alpha \in I}$.
$\\$ $\therefore Hom(\oplus_{\alpha \in I} G_\alpha, H) \cong \prod_{\alpha \in I} Hom(G_\alpha, H)$.