By contrapositive. Assume
ϕ is not
injective. Then exists some
a=b∈G such that
ϕ(a)=ϕ(b). Then,
eH=ϕ(eG)=ϕ(aa−1)=ϕ(a)ϕ(a−1)=ϕ(b)ϕ(a−1)=ϕ(ba−1)
Thus
ba−1∈ker(ϕ). Also, since
b=a, then
ba−1=eG pf. Thus
ker(ϕ)={eG}.
Lemma. If
b=a then
ba−1=e.
Proof. By contrapositive. If
ba−1=e then multiplying both sides by
a gives
b=a.