Question

prove that any two right cosets are either identical or disjoint? ​

Answer 1

Answer:

for a subgroup H of a group G is any two right or left cosets of H in G are identical or disjoint

Answer 2

Answer:

Any two right cosets are either identical or disjoint is proved.

Step-by-step explanation:

To prove: Two right cosets are either identical or disjoint.

Let H be a subgroup of G.

Also let Ha and Hb be two right cosets of H of G such that a, b ∈ G.

Case1. If Ha ∩ Hb = ∅.

Then, there is nothing to prove.

Thus, the two right cosets are disjoint.

Case2. If Ha ∩ Hb ≠ ∅.

To show: Ha = Hb

Since Ha ∩ Hb ≠ ∅ then there exists at least one element t ∈ Ha ∩ Hb.

⇒ t ∈ Ha and t ∈ Hb

⇒ t = $h_1a$ for some $h_1$ ∈ H and t = $h_2b$ for some $h_2$ ∈ H

⇒ $h_1a=h_2b$

Pre multiply both the sides by $h_1^{-1}$ as follows:

⇒ $h_1^{-1}(h_1a)=h_1^{-1}(h_2b)$

Using associativity for multiplication, we get

⇒ $(h_1^{-1}h_1)a=(h_1^{-1}h_2)b$

⇒ $ea=h_3b$, where $h_3=h_1^{-1}h_2\epsilon H$ and $h_1^{-1}h_1=e$ and $e$ is the identity element in H

⇒ $a=h_3b$

Now,

Pre multiply both the sides by the set H (Multiplication is possible only when the sets are finite)

⇒ $Ha=H(h_3b)$

         $=(Hh_3)b$

         $=Hb$ (Since $h_3$ ∈ H, so $Hh_2=H$)

⇒ Ha = Hb

This implies if Ha ∩ Hb ≠ ∅, then Ha = Hb.

Thus, the two right cosets are identical.

Hence proved.

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