Given that,"G" is finite group and "H" is subgroup of "G"
Claim: If "H\\triangleleft G" then, for any "a\\in G" ,"aH=Ha"
Proof:
Suppose, "H\\triangleleft G\\implies aHa^{-1}\\subset H"
Note that "H=a(a^{-1}H(a^{-1})^{-1})a^{-1}=a(a^{-1}Ha)a^{-1}\\subset aHa^{-1}"
Hence, from above two equation "aHa^{-1}=H\\implies aH=Ha"
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