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A subgroup [ilmath](H,\times_H:H\times H\rightarrow H[/ilmath] of a Group [ilmath](G,\times_G:G\times G\rightarrow G)[/ilmath] is a set [ilmath]H\subseteq G[/ilmath] which is a group under the operation [ilmath]\times_G[/ilmath] restricted to [ilmath]H\times H[/ilmath].


Given a group [ilmath](G,\times_G:G\times G\rightarrow G)[/ilmath] we say [ilmath](H,\times_H:H\times H\rightarrow H)[/ilmath] is a subgroup of [ilmath](G,\times_G)[/ilmath] if:

  1. [math]H\subset G[/math]
  2. the function [math]\times_H:H\times H\rightarrow G[/math] given by [math]\times_H(x,y)\mapsto\times_G(x,y)[/math] has [math]\text{Range}(\times_H)\subseteq H[/math]
    • That is to say it is closed. [math]\forall x\in H\forall y\in H[\times_H(x,y)\in H][/math]
  3. There exists an identity element [math]\in H[/math].
    • That is to say [math]\exists e\in H\forall x\in H[ex=xe=x][/math] where [math]xy[/math] denotes [math]\times_H(x,y)[/math]
  4. Every element has an inverse [math]\in H[/math]
    • That is to say [math]\forall x\in H\exists y\in H[xy=yx=e][/math]
  5. The operation is associative
    • That is to say [math]\forall x\in H\forall y\in H\forall z\in H[x(yz)=(xy)z][/math]

Just like a group

This makes it sound a lot harder than it really is.


Even numbers

Take the group [math](\mathbb{Z},+)[/math] and define [math]H=\{z\in\mathbb{Z}|z\text{ is even}\}[/math] then we have [math]H\subset\mathbb{Z}[/math] and we must check it is a group.

  1. It is closed under [math]+[/math] restricted to [math]H[/math] - an even + an even = even. (proof [math]2n+2m=2(m+n)[/math] and anything multiplied by 2 is even)
  2. The identity [ilmath]0\in H[/ilmath] - so we have that.
  3. Given an [math]x\in H[/math] we can see easily that the inverse, [math]-x[/math] is also even and thus [math]\in H[/math]
  4. Associativity is inherited

See also