Let \( H \) be a
subgroup of \( G \), and let \( e _ H, e _ G \) be their identities respectively, then
\[
e _ H = e _ G
\]
Since \( e _ H \) is the identity in \( H \) that means for any other \( h \in H \) we have that \( e _ H \cdot h = h \), but then \( e _ H \in H \) so that \( e _ H \cdot e _ H = e _ H \). On that same thought \( e _ H \in H \subseteq G \) so that \( e _ H \in G \) and thus \( e _ H \cdot e _ G = e _ G \) were we've used the fact that \( e _ G \) is the identity in \( G \). So we have the combined equality
\[
e _ H e _ H = e _ H e _ G \iff \left( e _ H ^ { -1 } \right) e _ H e _ H = \left( e _ H ^ { -1 } \right) e _ H e _ G \iff e _ H = e _ G
\]
So, in fact their identities are the same.