Third isomorphism theorem

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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 is about an isomorphism theorem in group theory.
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Suppose G is a group, and H and K are Normal subgroup (?)s of G, such that H \le K. Then we have the following natural isomorphism:

(G/H)/(K/H) \cong G/K

Where the isomorphism sends a coset Hg in G to the coset Kg in G.

Note that this statement makes sense at the level of a group isomorphism only when both H and K are normal in G. Otherwise, the statement is still true at the level of sets, but we cannot make sense of it as a group isomorphism.

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Other isomorphism theorems

Facts used


Given: A group G, with normal subgroups H and K, such that H \le K

To prove: (G/H)/(K/H) \cong G/K

Proof: Note first that all the three expressions for quotient groups make sense. G/H and G/K make sense because H,K are normal in G. Moreover, since normality satisfies intermediate subgroup condition, H is also normal in K.

Next, observe that K/H is a normal subgroup in G/H, because normality is image-closed: under the quotient map by H, the normal subgroup K of G gets sent to a normal subgroup K/H of G/H. Thus, the left side makes sense.

Let's now describe the isomorphism from the left side to the right side:

\psi: G/H \to G/K, \qquad \psi(gH) = gK

In other words, the map takes a coset of H and gives the corresponding coset of K. This is well-defined, because if h \in H, then h \in K, so (gh)K = g(hK) = gK.

Further, the map is a homomorphism. For this, observe that it sends the identity element to the identity element, preserves the group multiplication, and preserves the inverse map.

Further, the map is surjective, because any coset gK occurs as the image of gH under \psi.

Finally, we need to determine the kernel of the map. This is given by the set of gH such that gK = K. This is precisely those cosets of H that are in K, which is the same as the coset space K/H. Hence, the kernel of the map is precisely G/K.

Thus, the surjective homomorphism \psi:G/H \to G/K has kernel precisely K/H. By the first isomorphism theorem, we get:

(G/H)/(K/H) \cong G/K.


Textbook references

  • Abstract Algebra by David S. Dummit and Richard M. Foote, 10-digit ISBN 0471433349, 13-digit ISBN 978-0471433347, More info, Theorem 19, Section 3,3, Page 98
  • Algebra by Michael Artin, ISBN 0130047635, 13-digit ISBN 978-0130047632, More info, Exercise 8, Miscellaneous Problems, Page 236