Question

Let G be a group and H≤G.Prove that the only right coset of H in G that is a subring of G is H itself.?

Answer 1

Assuming the question (as clarified by comments) is:

Let `G` be a group and `H \leq G`. Prove that the only right coset of `H` in `G` that is a subgroup of `G` is `H` itself.

Explanation:

Let `G` be a group and `H \leq G`. For an element `g \in G`, the right coset of `H` in `G` is defined as:

`=>Hg = {hg : h \in H}`

Let us assume that `Hg \leq G`. Then the identity element `e \in Hg`. However, we know necessarily that `e \in H`.

Since `H` is a right coset and two right cosets must either be identical or disjoint, we can conclude `H = Hg` ∎

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In case this isn't clear, let's try a proof eliminating symbols.

Let `G` be a group and let `H` be a subgroup of `G`. For an element `g` belonging to `G`, call `Hg` the right coset of `H` in `G`.

Let us assume that the right coset `Hg` is a subgroup of `G`. Then the identity element `e` belongs to `Hg`. However, we already know that the identity element `e` belongs to `H`.

Two right cosets must either be identical or disjoint. Since `H` is a right coset, `Hg` is a right coset, and both contain `e`, they cannot be disjoint. Hence, `H` and `Hg` must be identical, or `H = Hg` ∎