Second isomorphism theorem

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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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 $H,N\leq G$ are subgroups and $N$ is normal in $G$, then $H\cap N$ is normal in $H$.
• Normality satisfies intermediate subgroup condition: If $H\leq K\leq G$ 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 $H,N\leq G$ 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 $h_{1}n_{1}h_{2}n_{2}\in HN$	Observe that $n_{1}h_{2}=h_{2}n_{3}$ for some $n_{3}\in N$, by the condition, so $h_{1}n_{1}h_{2}n_{2}=h_{1}h_{2}n_{3}n_{2}\in HN$.
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}