Problem 2.2.15

We had problems explaining to you why this problem makes sense, so here’s a write-up. In the definition of subgroup, the identity element in H is required to be the identity of G. One might require only that H have an identity element, not that it is the same as the identity in G. Show that if H has an identity at all, then it is the identity in G , so this definition would be equivalent to the one given.


Let e \in G be the identity element of the group G;
let e_{H} \in H be an element that behaves like the identity for for H;

Take any c \in H (since H \subset G, we also have c \in G).
We know that e_{H}c = ce_{H} = c since c \in H,
and also that ec = ce = c, since c \in G.

Equating the two, e_{H}c = ec, and a right-multiplication by c^{-1} gives e_{H} = e, 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 H, but works differently for some c \not\in H!
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 c \in H, we can just take e_{H} \in H. We get that e_{H}e_{H} = e_{H}. Now multiply by e_{H}^{-1} to conclude e_{H} = e.
The argument, in plain words: a faux-identity cannot exist, because it would square to itself, and only the actual identity does that. — Z