Question

Let G = {1,a,a^2,a^3}(a^4 = 1) be a group and H = {1, a^2} is a subgroup of G under multiplication.
Which are disjoint cosets.​

Answer 1

⟨G,⋆⟩ is a finite group and H is a finite subgroup of G. For a∈G, define aH={ah∣h∈H}

i.e. aH is a left coset.

a.

The order of a coset is defined as its cardinality.

any 2 sets have the same cardinality if and only if there is a bijection between them.

So to show any two left cosets have the same cardinality, it suffices to demonstrate a bijection between them.

Let aH and bH are 2 left cosets of H in G such that

aH={ah∣h∈H} and bH={bh∣h∈H} where a,b∈G

Let f be a mapping defined from aH to bH such that

f(ah)=bh ∀ h∈H

Let ah1=ah2 where h1,h2∈H

⟹a−1ah1=a−1ah2

⟹h1=h2

⟹bh1=bh2

⟹f(ah1)=f(ah2)

∴ f is well defined.

Let f(ah1)=f(ah2) where h1,h2∈H

⟹bh1=bh2

⟹b−1bh1=b−1bh2

⟹h1=h2

⟹ah1=ah2

∴ f is one-one mapping.

Let y∈bH then y=bh where h∈H

∃ah∈aH such that

f(ah)=hb=y

⟹y has a pre-image in aH

⟹ Every element of bH has a pre-image in aH

∴ f is an onto mapping.

∵ f is a well-defined one-one and onto mapping

⟹f is a bijective function

⟹  the Cosets aH and bH have the same cardinality.

∴ |aH|=|bH|

Step-by-step explanation: