Second isomorphism theorem

From Groupprops
Jump to: navigation, search
This article gives the statement, and possibly proof, of a basic fact in group theory.
View a complete list of basic facts in group theory
VIEW FACTS USING THIS: directly | directly or indirectly, upto two steps | directly or indirectly, upto three steps|
VIEW: Survey articles about this
This article is about an isomorphism theorem in group theory.
View a complete list of isomorphism theorems| Read a survey article about the isomorphism theorems

Name

This result is termed the second isomorphism theorem or the diamond isomorphism theorem (the latter name arises because of the diamond-like shape that can be used to describe the theorem).

Statement

General statement

Suppose G is a group, and H,N are two subgroups of G such that H normalizes N; in other words hN = Nh for every h \in H. Then, HN is a group, with N \underline{\triangleleft} HN (i.e., N is a normal subgroup of HN), and H \cap N \underline{\triangleleft} H (i.e. H \cap N is a normal subgroup of H), and:

HN/N \cong H/(H \cap N)

Particular case

Suppose G is a group, N is a normal subgroup, and H an arbitrary subgroup, such that HN = G. Then:

G/N \cong H/(H \cap N)

Secondisomorphismtheorem.png

Definitions used

Term Relevant definitions(s)
normal subgroup A subgroup A of a group B is termed a normal subgroup if for any a \in A and b \in B, bab^{-1} \in A. Equivalently, A is normal in B if and only if for any b \in B, bAb^{-1} = A.
isomorphism of groups Given groups A and B a map \varphi:A \to B is termed an isomorphism of groups if \varphi is bijective and \varphi(gh) = \varphi(g)\varphi(h) for all g,h \in A.
quotient group For a normal subgroup A of a group B, the quotient group B/A is the group whose elements are the cosets bA, b \in B, and where the product of two cosets is the unique coset comprising the products of elements in the two cosets.

Particular cases

Specific possibilities for the relationship between H and N

Note that for each of these, if we have HN = G, then the statement made for HN applies to G.

Situation Interpretation
N is trivial In this case, HN/N and H/(H \cap N) are actually the same thing, the isomorphism is the identity map, and the result gives no information.
H is trivial In this case, HN/N and H/(H \cap N) are both trivial groups with the trivial isomorphism between them. Again, the result gives no information.
H \cap N is trivial In this case, HN is an internal semidirect product with normal piece N and complement (and hence retract) H. The result says that HN/N \cong H, i.e., complement to normal subgroup is isomorphic to quotient.
H \cap N is trivial, both H and N normalize each other In this case, HN is an internal direct product of H and N. We get HN/N \cong H and HN/H \cong N.

Specific possibilities for G and the location of H and N within

Nature of G Nature/parameters for H and N Interpretation of H \cap N and HN Interpretation of H/(H \cap N) and HN/N Fact in this context that corresponds to second isomorphism theorem
group of integers \mathbb{Z} H = h\mathbb{Z}, N = n\mathbb{Z} If d = \operatorname{gcd}(h,n) and l = \operatorname{lcm}(h,n), then HN = d\mathbb{Z} and H \cap N = l\mathbb{Z} H/(H \cap N) = h\mathbb{Z}/d\mathbb{Z} which is cyclic of order h/d. HN/N = l\mathbb{Z}/n\mathbb{Z} which is cyclic of order l/n The arithmetic fact that hn = ld, i.e., the product of two numbers equals the product of their gcd and lcm.
vector space V of dimension v H is a subspace of dimension h, N is a subspace of dimension n H \cap N has dimension m and HN has dimension M; these dimensions are not determined by h,n The dimension of H/(H \cap N) is h - m and the dimension of HN/N is M - n The isomorphism between these (which turns out to also be a vector space isomorphism) preserves dimension, and we get h - m = M - n or h + n = M + m. This is a vector space analogue of the inclusion-exclusion principle.
G is an external direct product of groups C_i, i \in I There are subsets J,K of I such that H is the subgroup where all coordinates outside J are the identity, and N is the subgroup where all coordinates outside K are the identity. H \cap N corresponds to all coordinates outside J \cap K being the identity, and HN corresponds to J \cup K. H/(H \cap N) corresponds to (in a loose subgroup-quotient identification) J \setminus (J \cap K) and HN/N to (J \cup K) \setminus K. The fact that J \setminus (J \cap K) = (J \cup K) \setminus K. In finite cardinality terms, |J \cap K| + |J \cup K| = |J| + |K|.

Related facts

Facts about normal subgroups

General version of the result

  • Product formula: The set-theoretic version of the product formula establishes a bijection which is the same as the bijection of the second isomorphism theorem, but without the conditions of normality. The bijection is purely at the set-theoretic level.

Facts used

  1. Subgroup containment implies coset containment

Proof

Given: A group G, subgroups H, N \le G such that hN = Nh for all h \in H.

To prove: HN is a group, N is normal in HN, H \cap N is normal in H, and HN/N \cong H/(H \cap N).

Proof:

HN is a subgroup of G

Condition In symbols Justification
Closure under multiplication For h_1, h_2 \in H, n_1, n_2 \in N, want to show h_1n_1h_2n_2 \in HN Observe that n_1h_2 = h_2 n_3 for some n_3 \in N, by the condition, so h_1n_1h_2n_2 = h_1h_2n_3n_2 \in HN.
Identity element The identity element of G is in HN The identity element is in both H and N, so it is in HN.
Inverses Given h \in H, n \in N, (hn)^{-1} \in HN (hn)^{-1} = n^{-1}h^{-1} \in Nh^{-1} = h^{-1}N \subseteq HN. Thus, the inverse of any element in HN is also in HN.

N is normal in HN

Observe that, by the condition, hNh^{-1} = N for any h \in H. Thus, for h \in H and n \in N, we have hnN(hn)^{-1} = h(nNn^{-1})h^{-1} = hNh^{-1} = N. Thus, N is normal in HN.

H \cap N is normal in H

We now prove that H \cap N is normal in H. Pick h \in H, x \in H \cap N. Then hxh^{-1} \in H such H is a subgroup. Also, since hNh^{-1} =N for any h \in H, we have hxh^{-1} \in N, so hxh^{-1} \in H \cap N. Thus, h(H \cap N)h^{-1} \subseteq H \cap N, and we get that H \cap N is normal in H.

Definition of isomorphism and proof that it works

Finally, define the isomorphism \varphi between H/(H \cap N) and \! HN/N as follows:

\varphi(g(H \cap N)) = gN.

We check that this satisfies all the required conditions:

Condition Verification
Sends cosets to cosets Note first that if two elements are in the same coset of H \cap N, they are in the same coset of N. Thus, the map sends cosets of H \cap N to cosets of N. (This is fact (1)).
Well-defined with specified domain and co-domain Further, if g \in H, then gN \subseteq HN. In other words, if the original coset is in H, the new coset is in HN. Thus, the map \varphi is well-defined from H/(H \cap N) to \! HN/N.
Injective Suppose \varphi(a(H \cap N)) = \varphi(b(H \cap N)). That means that aN = bN, forcing a^{-1}b \in N. But we anyway have a,b \in H, so a^{-1}b \in H \cap N, forcing that a and b are in the same coset of H \cap N. Thus, a(H \cap N) = b(H \cap N).
Surjective Any left coset of N in HN can be written as gN where g \in HN. Thus, we can write g = ab where a \in H, b \in N. Then, gN = abN = a(bN) = aN, with a \in H. Thus, gN = \varphi(a(H \cap N)).
Homomorphism Suppose a,b \in H. Then, \varphi(a(H \cap N)b(H \cap N)) = \varphi(ab(H \cap N)) = abN = (aN)(bN) = \varphi(a(H \cap N))(\varphi(b(H \cap N)). Thus, the map is a homomorphism.

References

Textbook references

  • Abstract Algebra by David S. Dummit and Richard M. Foote, 10-digit ISBN 0471433349, 13-digit ISBN 978-0471433347, Page 97, Theorem 18 of Section 3.3, More info
  • Algebra by Michael Artin, ISBN 0130047635, 13-digit ISBN 978-0130047632, Page 236, Exercise 7 of Miscellaneous Problems, More info