Interpretation of First Isomorphism Theorem

Question

I understand that the first isomorphism theorem basically says that any homomorphism f from group G to group G' induces an isomorphism from G/Ker(f) to the f(G).

What I don't quite understand is the following passage from Herstein, "By the first isomorphism theorem for any normal subgroup N of G, G/N is a homomorphic image of G" which I do understand, but then he continues, "Thus, there is a one-to-one correspondence between homomorphic images of G and the normal subgroups of G." Is Herstein saying that the group G' is the homomorphic image of a given group G iff G' is isomorphic to G/N for some normal subgroup of G?

Thank you,

Matt

Answer 1

If I understand correctly, are are asking about the following.

You can think of any normal subgroup as a kernel of some homomorphism, and vice versa. It is obvious that any kernel is a normal subgroup. From the other hand, if $N<G$ is a normal subgroup, then it is the kernel of the canonical homomorphism $G\to G/N$ sendeing $g\mapsto gN$. It is straighforward to check that since $N$ was normal, the cosets $G/N$ indeed form a group under the operation $g_1N\cdot g_2N=g_1g_2N$.

Answer 2

I am reading "Topics in Algebra 2nd Edition" by I. N. Herstein.

Theorem 2.7.1 is the first isomorphism theorem.
Lemma 2.7.1 says if $N$ is a normal subgroup of $G$, then $G\ni x\mapsto Nx\in G/N$ is a surjective homomorphism. (i.e. $G/N$ is a homomorphic image of $G$.)

The author wrote:

Theorem 2.7.1 is important, for it tells us precisely what groups can be expected to arise as homomorphic images of a given group. These must be expressible in the form $G/K$, where $K$ is normal in $G$. But, by Lemma 2.7.1, for any normal subgroup $N$ of $G$, $G/N$ is a homomorphic image of $G$. Thus there is a one-to-one correspondence between homomorphic images of $G$ and normal subgroups of $G$.

Let $S$ be the set of all normal subgroups of $G$.
Let $T$ be the set of all homomorphic images of $G$. (two different elements of $T$ are not isomorphic.) By Lemma 2.7.1, for any $N\in S$, $G/N\in T$.
I think the author is saying $f: S\ni N\mapsto G/N\in T$ is bijective.
And I think the author is not correct.

By Theorem 2.7.1, if $\overline{G}\in T$, then $\overline{G}\cong G/K$ for some normal subgroup $K$ of $G$.
So, $f$ is surjective.

But, $f$ is not injective in general.
Let $D_4$ be a dihedral group.
Let $H_1:=\{\operatorname{id}, (1,3)(2,4), (1,4)(2,3), (1,2)(3,4)\}$.
Let $H_2:=\{\operatorname{id}, (2,4), (1,3), (1,3)(2,4)\}$.
Let $H_3:=\{\operatorname{id}, (1,3)(2,4), (1,4,3,2), (1,2,3,4)\}$.
Then, $H_1$ and $H_2$ and $H_3$ are subgroups of $G$.
Since $(D_4:H_1)=(D_4:H_2)=(D_4:H_3)=2$, $H_1$ and $H_2$ and $H_3$ are normal subgroups of $D_4$.
Since $o(D_4/H_1)=o(D_4/H_2)=o(D_4/H_3)=2$, $D_4/H_1\cong D_4/H_2\cong D_4/H_3$.
$H_1,H_2\in S$ and $H_1\neq H_2$, but $D_4/H_1=D_4/H_2\in T$.
So, $f$ is not injective in this case.