A description/proof of that subgroup of Abelian additive group is retract of group iff there is another subgroup such that group is sum of subgroups
Topics
About: group
The table of contents of this article
Starting Context
- The reader knows a definition of group.
- The reader knows a definition of subgroup.
- The reader knows a definition of map.
Target Context
- The reader will have a description and a proof of the proposition that for any Abelian additive group, any subgroup is a retract of the group if and only if there is another subgroup such that the group is the sum of the subgroups.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Description
For any Abelian additive group, \(G\), any subgroup, \(G_1 \subseteq G\), is a retract of \(G\) if and only if there is a subgroup, \(G_2 \subseteq G\), such that \(G_1 \cap G_2 = \{0\}\) and \(G_1 + G_2 = G\), which means that any element, \(g \in G\), is uniquely \(g = g_1 + g_2\) where \(g_1 \in G_1\) and \(g_2 \in G_2\).
2: Proof
Let us suppose that \(G_1 \cap G_2 = \{0\}\) and \(G_1 + G_2 = G\). Let us define a map, \(f: G \rightarrow G_1, g = g_1 + g_2 \mapsto g_1\). Then, \(f\vert_{G_1}: g_1 \mapsto g_1\), an identical map. Is \(f\) a group homomorphism? For any \(g, h \in G\), \(g = g_1 + g_2\) and \(h = h_1 + h_2\), \(f (g + h) = f (g_1 + g_2 + h_1 + h_2) = f (g_1 + h_1 + g_2 + h_2) = g_1 + h_1 = f (g_1 + g_2) + f (h_1 + h_2) = f (g) + f (h)\), so, yes. So, \(f\) is a retraction and \(G_1\) is a retract of \(G\).
Let us suppose that \(G_1\) is a retract of \(G\) by a group homomorphism, \(f: G \rightarrow G_1\). Let us define \(G_2\) as \(\{g_2 \in G\vert f (g_2) = 0\}\). Is \(G_2\) a group? For any \(g_2 \in G_2\), \(0 = f (0) = f (g_2 - g_2) = f (g_2) + f (- g_2) = f (- g_2)\) where \(f (0) = 0\) because \(0 \in G_1\), so, \(- g_2 \in G_2\). For any \(g_2, h_2 \in G_2\), \(f (g_2 + h_2) = f (g_2) + f (h_2) = 0\), so, \(g_2 + h_2 \in G_2\). So, \(G_2\) is indeed a group. \(G = G_1 + G_2\)? For any \(g \in G\), \(g = f (g) + g - f (g)\), \(f (g - f (g)) = f (g) + f (- f (g)) = f (g) - f (g) = 0\) where \(f (- f (g)) = - f (g)\) because \(- f (g) \in G_1\), so, \(g = g_1 + g_2\) where \(g_1 = f (g) \in G_1\) and \(g_2 = g - f (g) \in G_2\). Is the division unique? Let us suppose that \(g = g_1 + g_2 = h_1 + h_2\). \(f (g) = f (g_1) + f (g_2) = g_1 = f (h_1) + f (h_2) = h_1\); \(g_2 = g - g_1 = g - h_1 = h_2\), so, unique. \(G_1 \cap G_2 = \{0\}\), because for any \(0 \neq g_1 \in G_1\), \(f (g_1) = g_1 \neq 0\), so, \(g_1 \notin G_2\).
