In the definition of subgroup, the identity element in is required to be the identity of . One might require only that have an identity element, not that it is the same as the identity in . Show that if has an identity at all, then it is the identity in , so this definition would be equivalent to the one given.
Let be the identity element of the group ;
let be an element that behaves like the identity for for ;
Take any (since , we also have ).
We know that since ,
and also that , since .
Equating the two, , and a right-multiplication by gives , Qed.
The argument could also be: since the identity element is unique in any group, and since a subgroup consists only of the elements of the “parent” group, the identity elements of the two groups must coincide. — M
One has to be careful with that second argument! Sure, there is only one identity element in every group, but that doesn’t mean (in itself) that there’s can’t be a faux-identity, which works like the identity for elements of a subgroup , but works differently for some !
In fact, faux-identities can’t exist, and the argument above establishes that.
If we read the proof carefully, we’ll realise that it can be simplified even further! Instead of considering an arbitrary , we can just take . We get that . Now multiply by to conclude .
The argument, in plain words: a faux-identity cannot exist, because it would square to itself, and only the actual identity does that. — Z