Suppose

\(\displaystyle{a}{H}\cap{b}{H}\ne\emptyset\)

Let \(\displaystyle{x}\in{a}{H}\cap{b}{H}\), then there exist \(\displaystyle{h}_{{1}},{h}_{{2}}\in{H}\)

Such that,

\(\displaystyle{x}={a}{h}_{{1}}&{x}={b}{h}_{{2}}\)

\(\displaystyle{a}={x}{{h}_{{1}}^{{-{{1}}}}}&{b}={x}{{h}_{{2}}^{{-{{1}}}}}\)

\(\displaystyle{a}={b}{h}_{{2}}{{h}_{{1}}^{{-{{1}}}}}\)

\(\displaystyle{a}{H}={b}{h}_{{2}}{{h}_{{1}}^{{-{{1}}}}}{H}={b}{H}\)

(since \(\displaystyle{a}{H}={H}{\quad\text{if}\quad}{a}\in{H}\))

Now, if \(\displaystyle{a}\in{b}{H}\), using above theorem

Since \(\displaystyle{b}\in{b}{H}\), then \(\displaystyle{a}{H}={b}{H}\Rightarrow{b}\in{a}{H}\)

\(\displaystyle{a}\in{H}{\quad\text{and}\quad}{a}\in{a}{H}\Rightarrow{a}\in{a}{H}\cap{b}{H}\)

\(\displaystyle\Rightarrow{a}{H}\cap{b}{H}\ne{0}\)

\(\displaystyle\Rightarrow{S}{o}{a}{H}\cap{b}{H}={0}\)

\(\displaystyle\Rightarrow{b}\in{a}{H}\)

Hence G be a group, H a subgroup of G, and a and b elements of G. If \(\displaystyle{a}\in{b}{H}\) then \(\displaystyle{b}\in{a}{H}\).

\(\displaystyle{a}{H}\cap{b}{H}\ne\emptyset\)

Let \(\displaystyle{x}\in{a}{H}\cap{b}{H}\), then there exist \(\displaystyle{h}_{{1}},{h}_{{2}}\in{H}\)

Such that,

\(\displaystyle{x}={a}{h}_{{1}}&{x}={b}{h}_{{2}}\)

\(\displaystyle{a}={x}{{h}_{{1}}^{{-{{1}}}}}&{b}={x}{{h}_{{2}}^{{-{{1}}}}}\)

\(\displaystyle{a}={b}{h}_{{2}}{{h}_{{1}}^{{-{{1}}}}}\)

\(\displaystyle{a}{H}={b}{h}_{{2}}{{h}_{{1}}^{{-{{1}}}}}{H}={b}{H}\)

(since \(\displaystyle{a}{H}={H}{\quad\text{if}\quad}{a}\in{H}\))

Now, if \(\displaystyle{a}\in{b}{H}\), using above theorem

Since \(\displaystyle{b}\in{b}{H}\), then \(\displaystyle{a}{H}={b}{H}\Rightarrow{b}\in{a}{H}\)

\(\displaystyle{a}\in{H}{\quad\text{and}\quad}{a}\in{a}{H}\Rightarrow{a}\in{a}{H}\cap{b}{H}\)

\(\displaystyle\Rightarrow{a}{H}\cap{b}{H}\ne{0}\)

\(\displaystyle\Rightarrow{S}{o}{a}{H}\cap{b}{H}={0}\)

\(\displaystyle\Rightarrow{b}\in{a}{H}\)

Hence G be a group, H a subgroup of G, and a and b elements of G. If \(\displaystyle{a}\in{b}{H}\) then \(\displaystyle{b}\in{a}{H}\).