Groups, Correspondence Theorems
Correspondence Theorems
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The correspondence theorem, or isomorphism theorem,
is sometimes presented as three separate theorems.
In fact they are often called the first second and third isomorphism theorems.
Well I couldn't remember where to draw the line,
so I just clumped them all together.
Hope that's ok.
This theorem, or theorems if you prefer,
asserts the equivalence of the subgroups or normal subgroups of G
with those in the factor group H.
In other words, subgroups containing K, or normal subgroups containing K,
correspond 1-1 with subgroups or normal subgroups in the factor group G/K.
Let's get started.
Let K be a normal subgroup of G, with factor group H.
If R is a subgroup of G, its image
S is a subgroup of H.
Conversely, if S is a subgroup of H, its preimage R is a subgroup of G.
Use the
a/b criterion to verify this.
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Applying the
x*w/x criterion,
verify R is normal in G iff its image is normal in H,
and S is normal in H iff its preimage is normal in G.
Assume H is normal in G, and K is normal in both H and G.
Let Q be the factor group G/K.
By the above, the image of H in Q is a normal subgroup of Q.
This leads to a factor group R.
We will show that R is the same as G/H.
The map is straightforward,
though somewhat technical.
Given a coset of H in G,
choose a cosrep x,
let it define a coset of K in G, i.e. an element of Q,
and let that element of Q define a coset of the image of H in Q,
i.e. an element of R.
We need to show this is well defined.
Multiply x by y for some y in H, and see if this changes anything.
Moving to Q, the image of x is multiplied by something in the image of H.
This leads to the same element of R.
So it doesn't matter which cosrep we use.
Use the property of group homomorphism, from G to Q to R,
to show this map commutes with *, and is a group homomorphism.
Since Q is the image of G and R is the image of Q, the map is onto.
Finally, different cosets of H in G lead to
different cosets of the image of H in Q, hence the map is an isomorphism.
If G is finite,
the index of K in G (|Q|),
is equal to the index of K in H (the size of the image of H in Q)
times the index of H in G (the size of R).