**Proof:** Let HK be a subgroup. Then (HK)^{-1} = HK

⟹ K^{-1} H^{-1} = HK

⟹ KH = HK [ Since H and K being subgroups H^{-1} = H and K^{-1} = K]

# Conversely let HK = KH.

To prove that HK is a subgroup, we are to show that (HK) (HK)

^{-1}= HK.

Now, (HK) (HK)^{-1} = (HK) (H^{-1} k^{-1})

= H(KK^{‑1})H^{-1} [by associatively]

=(HK)H^{-1} [since K is a subgroup, KK^{‑1}]^{ }

= (KH)H^{-1} [since HK = KH]

= K(HH^{-1}) [since H is a subgroup, HH^{‑1} = H]

= KH

= HK

*Hence HK is a subgroup. (Proved)
*