Second isomorphism theorem

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:

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

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

• Normality satisfies transfer condition: If $H, N \le G$ are subgroups and $N$ is normal in $G$, then $H \cap N$ is normal in $H$.
• Normality satisfies intermediate subgroup condition: If $H \le K \le G$ are groups and $H$ is normal in $G$, then $H$ is normal in $K$.

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}