EDIT: Whoops, the other poster is right. I did indeed not see the word "onto". It's not very common to see mathematicians use the word "onto" applied to a homomorphism in the literature. The more common term is "epimorphism", which doesn't mean the same thing as "surjection", but is provably the same thing in group theory. If you want to impress your lecturers, drop the word "epimorphism" into everyday conversation. So, the original proof follows: Because θ is surjective, for any h∈H, you can find a g∈G such that θ(g)=h. So for any h∈H and n∈N: h θ(n) h' = θ(g) θ(n) θ(g') = θ(gng') ∈ θ(N) It's also possible to prove this without referencing elements, but that's advanced stuff.
Answers & Comments
Verified answer
To prove KerΦ is a subgroup, we need to check that:
1) It is non-empty (typically by showing that the identity is in it)
2) It is closed under the group's operation
3) For any element of KerΦ, the inverse is also in KerΦ
1) is trivial, once you know that Φ(e) = e
2) Let x and y be in KerΦ. That is:
Φ(x) = e
Φ(y) = e
We need to show that xy is in KerΦ, that is:
Φ(xy) = e
But using the homomorphism property:
Φ(xy) = Φ(x)Φ(y) = ee = e
3) Suppose x is in KerΦ and y is the inverse. That is:
Φ(x) = e
xy = yx = e
We need to show that y is in KerΦ, that is:
Φ(y) = e
Now, consider:
Φ(xy) = Φ(yx) = Φ(e)
Φ(x)Φ(y) = Φ(y)Φ(x) = Φ(e)
Since x and e are both in the kernel, they will map to e, so:
eΦ(y) = Φ(y)e = e
Φ(y) = e
Therefore, y is also in the kernel.
These three parts total a proof that KerΦ is a subgroup of G.
EDIT: Whoops, the other poster is right. I did indeed not see the word "onto". It's not very common to see mathematicians use the word "onto" applied to a homomorphism in the literature. The more common term is "epimorphism", which doesn't mean the same thing as "surjection", but is provably the same thing in group theory. If you want to impress your lecturers, drop the word "epimorphism" into everyday conversation. So, the original proof follows: Because θ is surjective, for any h∈H, you can find a g∈G such that θ(g)=h. So for any h∈H and n∈N: h θ(n) h' = θ(g) θ(n) θ(g') = θ(gng') ∈ θ(N) It's also possible to prove this without referencing elements, but that's advanced stuff.