# Properties of Cosets

__Theorem 1__**:** If $$h \in H$$, then the right (or left) coset $$Hh$$ or $$hH$$ of $$H$$ is identical to $$H$$, and conversely.

**Proof:** Let $$h’$$ be an arbitrary element of $$H$$ so that $$hh’ \in hH$$.

Again, since $$H$$ is a subgroup, we have

$$h \in H,\,\,\,h’ \in H\, \Rightarrow \,hh’ \in H$$

Thus every element of $$hH$$ is also an element of $$H$$.

Hence \[hH \subset H\, — \left( i \right)\]

Again $$h’ = \left( {h{h^{ – 1}}} \right)h’ = h\left( {{h^{ – 1}}h’} \right) \in hH$$

This shows that every element of $$H$$ is also an element of $$hH$$.

Hence \[H \subset hH\,\, — \left( {ii} \right)\]

From (i) and (ii) it follows that $$hH = H$$

Similarly, we can show that $$Hh = H$$

Conversely, $$Hh = H \Rightarrow eH \in H \Rightarrow h \in H$$ and similarly $$hH \Rightarrow H \Rightarrow h \in H$$.

__Theorem 2__**:** Any two right (or left) cosets of $$H$$ are either disjoint or identical.

**Proof:** Let $$H$$ be a subgroup of a group $$G$$ and let $$aH$$ and $$bH$$ be two left cosets. Suppose these cosets are not disjoint. Then they possess an element, say $$c$$, in common. Then $$c$$ may be written as $$c = ah$$, and also as $$c = ah’$$, where $$h$$ and $$h’$$ are in $$H$$.

Therefore,

\[ah = bh’\]

\[a = bh'{h^{ – 1}}\]

Since $$H$$ is a subgroup, $$h'{h^{ – 1}} \in H$$.

Let $$h'{h^{ – 1}} = h”$$ then $$a = bh”$$. Hence $$aH = \left( {bh”} \right)H \Rightarrow aH = b\left( {h”H} \right) = bH$$

Therefore the two left cosets are identical if they are not disjoint.

Thus either $$aH \cap bH = \phi $$ or $$H = bH$$

A similar result can be shown to hold for right cosets.

__Theorem 3__**:** If $$H$$ is finite, the number of elements in a right (or left) coset of $$H$$ is equal to the order of $$H$$.

**Proof:** The mapping $$f:H \to Ha$$, defined by $$f\left( {{h_i}} \right) = {h_i}a$$, is obviously onto.

It is one-one, since

\[f\left( {{h_i}} \right) = f\left( {{h_j}} \right) \Rightarrow {h_i}a = {h_j}a\]

\[ \Rightarrow {h_i} = {h_j}\]

From the right cancellation law it is onto, since an element $$ha$$ belonging to $$Ha$$ is the $$f – $$ image of $$h$$ belonging to $$H$$.

It follows that the number of elements in a left coset of $$H$$ is the same as that of $$H$$.

Similarly the number of elements in a left coset of $$H$$ is the same as that of $$H$$.

umair

April 1@ 11:41 amhow f(hi) = f(h2)