# 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.

Proof

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