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'''[[Goursat]]'s lemma''' is an [[algebra]]ic [[theorem]] about [[subgroup]]s of the [[Direct product of groups|direct product]] of two [[Group (mathematics)|groups]].
 
It can be stated as follows.
:Let <math>G</math>, <math>G'</math> be groups, and let <math>H</math> be a subgroup of <math>G\times G'</math> such that the two [[projection (mathematics)|projections]] <math>p_1: H\rightarrow G</math> and <math>p_2: H\rightarrow G'</math> are [[surjective]] (i.e., <math>H</math> is a [[subdirect product]] of <math>G</math> and <math>G'</math>). Let <math>N</math> be the kernel of <math>p_2</math> and <math>N'</math> the [[Kernel (algebra)|kernel]] of <math>p_1</math>. One can identify <math>N</math> as a [[normal subgroup]] of <math>G</math>, and <math>N'</math> as a normal subgroup of <math>G'</math>. Then the image of <math>H</math> in <math>G/N\times G'/N'</math> is the [[graph of a function|graph]] of an [[isomorphism]] <math>G/N\approx G'/N'</math>.
 
An immediate consequence of this is that the subdirect product of two groups can be described as a [[Direct product of groups#Fiber products|fiber product]] and vice versa.
 
== Proof of Goursat's lemma ==
 
Before proceeding with the [[Mathematical proof|proof]], <math>N</math> and <math>N'</math> are shown to be normal in <math>G \times \{e'\}</math> and <math>\{e\} \times G'</math>, respectively. It is in this sense that <math>N</math> and <math>N'</math> can be identified as normal in ''G'' and ''G''', respectively.
 
Since <math>p_2</math> is a [[homomorphism]], its kernel ''N'' is normal in ''H''. Moreover, given <math>g \in G</math>, there exists <math>h=(g,g') \in H</math>, since <math>p_1</math> is surjective.  Therefore, <math>p_1(N)</math> is normal in ''G'', viz:
:<math>gp_1(N)=p_1(h)p_1(N)=p_1(hN)=p_1(Nh)=p_1(N)g</math>.
It follows that <math>N</math> is normal in <math>G \times \{e'\}</math> since
: <math>(g,e')N = (g,e')(p_1(N) \times \{e'\}) = gp_1(N) \times \{e'\} = p_1(N)g \times \{e'\} = (p_1(N) \times \{e'\})(g,e')=N(g,e')</math>.
 
The proof that <math>N'</math> is normal in <math>\{e\} \times G'</math> proceeds in a similar manner.
 
Given the identification of <math>G</math> with <math>G \times \{e'\}</math>, we can write <math>G/N</math> and <math>gN</math> instead of <math>(G \times \{e'\})/N</math> and <math>(g,e')N</math>, <math>g \in G</math>.  Similarly, we can write <math>G'/N'</math> and <math>g'N'</math>, <math>g' \in G'</math>.
 
On to the proof. Consider the map <math>H \rightarrow G/N \times G'/N'</math> defined by <math>(g,g') \mapsto (gN, g'N')</math>. The image of <math>H</math> under this map is <math>\{(gN,g'N') | (g,g') \in H \}</math>. This [[Relation (mathematics)|relation]] is the graph of a [[well-defined]] function <math>G/N \rightarrow G'/N'</math> provided <math>gN=N \Rightarrow g'N'=N'</math>, essentially an application of the [[vertical line test]].
 
Since <math>gN=N</math> (more properly, <math>(g,e')N=N</math>), we have <math>(g,e') \in N \subset H</math>. Thus <math>(e,g') = (g,g')(g^{-1},e') \in H</math>, whence <math>(e,g') \in N'</math>, that is, <math>g'N'=N'</math>. Note that by symmetry, it is immediately clear that <math>g'N'=N' \Rightarrow gN=N</math>, i.e., this function also passes the [[horizontal line test]], and is therefore [[injective function|one-to-one]]. The fact that this function is a surjective group homomorphism follows directly.
 
== References ==
 
* [[Ken Ribet|Kenneth A. Ribet]] (Autumn 1976), "[[Galois]] [[Group action|Action]] on Division Points of [[Abelian Variety|Abelian Varieties]] with Real Multiplications", ''[[American Journal of Mathematics]]'', Vol. 98, No. 3, 751–804.
 
[[Category:Algebra]]
[[Category:Lemmas]]
[[Category:Articles containing proofs]]

Latest revision as of 21:00, 26 March 2013

Goursat's lemma is an algebraic theorem about subgroups of the direct product of two groups.

It can be stated as follows.

Let G, G be groups, and let H be a subgroup of G×G such that the two projections p1:HG and p2:HG are surjective (i.e., H is a subdirect product of G and G). Let N be the kernel of p2 and N the kernel of p1. One can identify N as a normal subgroup of G, and N as a normal subgroup of G. Then the image of H in G/N×G/N is the graph of an isomorphism G/NG/N.

An immediate consequence of this is that the subdirect product of two groups can be described as a fiber product and vice versa.

Proof of Goursat's lemma

Before proceeding with the proof, N and N are shown to be normal in G×{e} and {e}×G, respectively. It is in this sense that N and N can be identified as normal in G and G', respectively.

Since p2 is a homomorphism, its kernel N is normal in H. Moreover, given gG, there exists h=(g,g)H, since p1 is surjective. Therefore, p1(N) is normal in G, viz:

gp1(N)=p1(h)p1(N)=p1(hN)=p1(Nh)=p1(N)g.

It follows that N is normal in G×{e} since

(g,e)N=(g,e)(p1(N)×{e})=gp1(N)×{e}=p1(N)g×{e}=(p1(N)×{e})(g,e)=N(g,e).

The proof that N is normal in {e}×G proceeds in a similar manner.

Given the identification of G with G×{e}, we can write G/N and gN instead of (G×{e})/N and (g,e)N, gG. Similarly, we can write G/N and gN, gG.

On to the proof. Consider the map HG/N×G/N defined by (g,g)(gN,gN). The image of H under this map is {(gN,gN)|(g,g)H}. This relation is the graph of a well-defined function G/NG/N provided gN=NgN=N, essentially an application of the vertical line test.

Since gN=N (more properly, (g,e)N=N), we have (g,e)NH. Thus (e,g)=(g,g)(g1,e)H, whence (e,g)N, that is, gN=N. Note that by symmetry, it is immediately clear that gN=NgN=N, i.e., this function also passes the horizontal line test, and is therefore one-to-one. The fact that this function is a surjective group homomorphism follows directly.

References