Normal subgroup equals kernel of homomorphism

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This article gives the statement, and possibly proof, of a basic fact in group theory.
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This article gives a proof/explanation of the equivalence of multiple definitions for the term normal subgroup
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Statement

Verbal statement

A subgroup of a group occurs as the Kernel (?) of a group homomorphism if and only if it is normal.

Symbolic statement

A subgroup N of a group G occurs as the kernel of a group homomorphism if and only if, for every g in G, gNg^{-1} \subseteq N.

Definitions used

Kernel of a group homomorphism

A map \varphi: G \to H is a homomorphism of groups if

  • \varphi(gh) = \varphi(g)\varphi(h) for all g, h in G
  • \varphi(e) = e
  • \varphi(g^{-1}) = (\varphi(g))^{-1}

The kernel of \varphi is defined as the inverse image of the identity element under \phi.

Normal subgroup

For the purpose of this statement, we use the following definition of normality: a subgroup H is normal in a group G if H contains each of its conjugate subgroups, that is, gNg^{-1} \subseteq N for every g in G.

Related facts

Closely related to this are the isomorphism theorems.

Proof

Kernel of homomorphism implies normal subgroup

Let \varphi: G \to H be a homomorphism of groups. We first prove that the kernel (which we call N) of \phi is a subgroup:

  • Identity element: Since \varphi(e) = e, e is contained in N
  • Product: Suppose a, b are in N. Then \varphi(a) = e and \varphi(b) = e. Using the fact that \varphi(ab) = \varphi(a)\varphi(b), we conclude that \varphi(ab) = e. Hence ab is also in N.
  • Inverse: Suppose a is in N. Then \varphi(a) = e. Using the fact that \varphi(a^{-1} = \varphi(a)^{-1}, we conclude that \varphi(a^{-1}) = e. Hence, a^{-1} is also in N.

Now we need to prove that N is normal. In other words, we must show that if g is in G and n is in N, then gng^{-1} is in N.

Since n is in N, \phi(n) = e.

Consider \varphi(gng^{-1}) = \varphi(g)\varphi(n)\varphi(g^{-1}) = \varphi(g)\varphi(g^{-1}) = \varphi(gg^{-1}) = \varphi(e) = e. Hence, gng^{-1} must belong to N.

Normal subgroup implies kernel of homomorphism

Let N be a normal subgroup of a group G. Then, N occurs as the kernel of a group homomorphism. This group homomorphism is the quotient map \varphi: G \to G/N, where G/N is the set of cosets of N in G.

The map is defined as follows:

\varphi(x) = xN

Notice that the map is a group homomorphism if we equip the coset space G/N with the following structure:

(aN)(bN)=abN

This gives a well-defined group structure because, on account of N being normal, the equivalence relation of being in the same coset of N yields a congruence.

Explicitly:

  1. The map is well-defined, because if a' = an_1, b' = bn_2 for n_1,n_2 \in N, then a'b' = an_1bn_2 = ab(b^{-1}n_1bn_2) \in abN (basically, we're using that bN = Nb).
  2. The image of the map can be thought of as a group because it satisfies associativity (((aN)(bN))(cN) = (aN)((bN)(cN))), has an identity element (N itself), has inverses (the inverse of aN is a^{-1}N)

Further information: quotient map

References

Textbook references

  • Abstract Algebra by David S. Dummit and Richard M. Foote, 10-digit ISBN 0471433349, 13-digit ISBN 978-0471433347, More info, Page 82, Proposition 7