249: Subgroup of Abelian Additive Group Is Retract of Group Iff There Is Another Subgroup Such That Group is Sum of Subgroups

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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

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Topics

About: group

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof

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Starting Context

• The reader knows a definition of group.
• The reader knows a definition of subgroup.
• The reader knows a definition of map.

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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.

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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.

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Main Body

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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 $g_2\in G_2$.

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2: Proof

Let us suppose that $G_1\cap G_2=\{0\}$ and $G_1+G_2=G$. Let us define a map, $f:G\to G_1,g=g_1+g_2\mapsto g_1$. Then, $f{|}_{G_1}:g_1\mapsto g_1$, 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, $f:G\to G_1$. Let us define $G_2$ as $\{g_2\in G|f(g_2)=0\}$. 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 $0\ne g_1\in G_1$, $f(g_1)=g_1\ne 0$, so, $g_1\notin G_2$.

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References

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