# Proof

(related to Proposition: Properties of Cosets)

In the following let $$(G,\ast)$$ be a group and $$H\subseteq G$$ its subgroup.

#### Proof of (1)

If $$|H|=n$$, then $$|Ha|\le n$$, which follows from the definition of a coset. But $$|Ha| \not < n$$. For if $$|Ha| < n$$, then there would be at least two different elements $$g,h\in H$$, $$g\neq h$$ and $$g\ast a=h\ast a$$. This is equivalent to $$g=h$$, which is a contradiction to $$g\neq h$$. Thus it must be $$|Ha|=|H|$$. Analogously, we can prove $$|aH|=|H|$$.

#### Proof of (2)

We have to prove that $$a\sim_l b:=a\in bH$$ (respectively $$a\sim_r b:=a\in Hb$$) define equivalence relations, by proving their reflexivity, symmetry and transitivity.

### Reflexivity

Because $$H\subset G$$ is a subgroup of $$G$$, the neutral element of $$G$$ is also the neutral element of $$H$$. Therefore, following the definition of cosets, $$a\in aH$$ and $$a\in Ha$$. So $$a\sim_l a$$ (respectively $$a\sim_r a$$) are reflexive.

### Symmetry

Assume $$a\sim_l b$$. From $$a\in bH$$ it follows that there exists $$x\in H$$ with $$a=b\ast x$$, which is equivalent to $$a\ast x^{-1}=b$$. Because also $$x^{-1}\in H$$, it follows that $$b\in aH$$, so $$b\sim_l a$$ is symmetric. Respectively, assume $$a\sim_r b$$. From $$a\in Hb$$ it follows that there exists $$x\in H$$ with $$a=x\ast b$$, which is equivalent to $$x^{-1}a=b$$. Because also $$x^{-1}\in H$$, it follows that $$b\in Ha$$, so $$b\sim_r a$$ is symmetric.

### Transitivity

Assume $$a\sim_l b$$ and $$b\sim_l c$$. From $$a\in bH$$ it follows that there exists $$x\in H$$ with $$a=b\ast x$$, which is equivalent to $$a\ast x^{-1}=b$$. From $$b\in cH$$ it follows that there exists $$y\in H$$ with $$b=c\ast y$$. From both results, it follows that $$a\ast x^{-1}=c\ast y$$, which is equivalent to $$a=c\ast y\ast x$$. Because $$H$$ is closed under $$\ast$$, $$y\ast x\in H$$. Therefore, $$a\sim_l c$$ is transitive. Respectively, assume $$a\sim_r b$$ and $$b\sim_r c$$. From $$a\in Hb$$ it follows that there exists $$x\in H$$ with $$a=x\ast b$$, which is equivalent to $$x^{-1}\ast a=b$$. From $$b\in Hc$$ it follows that there exists $$y\in H$$ with $$b=y\ast c$$. From both results, it follows that $$x^{-1}\ast a=y\ast c$$, which is equivalent to $$a=x\ast y\ast c$$. Because $$H$$ is closed under $$\ast$$, $$x\ast y\in H$$. Therefore, $$a\sim_r c$$ is transitive.

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### References

#### Bibliography

1. Ayres jr. Frank: "Theory and Problems of Modern Algebra", McGraw-Hill Book Company Europe, 1978