# Third isomorphism theorem

## Contents

This article gives the statement, and possibly proof, of a basic fact in group theory.
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## Statement

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.

## Proof

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:

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

## References

### 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