Mapping Elements in Quotient Group G/N to Isomorphism f:G/N--H

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In summary, the conversation discusses the possibility of an element in the quotient group G/N mapping back onto itself through an isomorphism f, even if it does not have a corresponding element in H. The example of the quotient group of symmetries of an n-gon, where n is even, is used to illustrate this concept. It is mentioned that the quotient group may contain every permutation in G, but this is incorrect due to rotations in G that do not map to H. The conversation also mentions an example using the dihedral group D_4 and its normal subgroup N, showing that the elements of D_4/N are essentially cosets of N.
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Phred Willard
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Can an element of a quotient group G/N in an isomorphism f:G/N--H map back onto itself if it does not have a corresponding element in H? The example I am looking at is the quotient group of the group of symmetries of an n-gon, G, where n is an even number and N equal to the normal subgroup containing the identity and a 180 rotation. H represents another n-gon with an even n, but is less than the order of G.

I have found that the quotient group G/N contains every permutation in G (which I think is incorrect) because for f to be an isomorphism, G contains rotations that don't map to H.

What's going on here?
 
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Phred Willard said:
Can an element of a quotient group G/N in an isomorphism f:G/N--H map back onto itself if it does not have a corresponding element in H? The example I am looking at is the quotient group of the group of symmetries of an n-gon, G, where n is an even number and N equal to the normal subgroup containing the identity and a 180 rotation. H represents another n-gon with an even n, but is less than the order of G.

I have found that the quotient group G/N contains every permutation in G (which I think is incorrect) because for f to be an isomorphism, G contains rotations that don't map to H.

What's going on here?

Are you referring to one of the isomorphism theorems?

Let's do a small example, with the dihedral group [itex]D_4[/itex] (the group of symmetries of a square). The 8 elements are [itex]\{ e, h, v, d_1, d_2, r_{90}, r_{180}, r_{270} \}[/itex]. If [itex]N=\{ e, r_{180} \}[/itex] then what are the elements of [itex]D_4/N[/itex]? (From Lagrange's theorem, we expect only 4 elements in this group.)

[itex]eN = r_{180}N = N[/itex]
[itex]vN = hN = \{ h, v \}[/itex]
[itex]d_1N = d_2N = \{ d_1, d_2 \}[/itex]
[itex]r_{90}N = r_{270}N = \{ r_{90}, r_{270} \}[/itex]

Notice that the elements of [itex]D_4/N[/itex] are essentially _cosets_ of N; so I am not sure what you mean by "G/N contains every permutation."
 

Related to Mapping Elements in Quotient Group G/N to Isomorphism f:G/N--H

1. What is the purpose of mapping elements in quotient group G/N to isomorphism f:G/N--H?

The purpose of this mapping is to establish a relationship between the quotient group G/N and another group H, where the elements in G/N are mapped to isomorphic elements in H. This allows us to compare and analyze the structures of the two groups.

2. How is the mapping f:G/N--H defined?

The mapping f:G/N--H is defined by assigning each element in G/N to an isomorphic element in H in a consistent and systematic manner. This can be done using various methods, such as defining a function or creating a table.

3. How do we determine if the mapping f:G/N--H is an isomorphism?

In order for the mapping f:G/N--H to be an isomorphism, it must satisfy two conditions: it must be a homomorphism, meaning that it preserves the group structure, and it must be bijective, meaning that it is both injective and surjective. We can check these conditions by verifying that the mapping preserves the group operation and that every element in H has a unique preimage in G/N.

4. What information can we learn from mapping elements in quotient group G/N to isomorphism f:G/N--H?

By mapping elements in G/N to isomorphic elements in H, we can determine the structure and properties of H based on the known structure and properties of G/N. This can be useful in understanding the relationship between two groups and in solving problems in abstract algebra.

5. Can the mapping f:G/N--H be used to prove the isomorphism between G/N and H?

Yes, if we can establish that the mapping f:G/N--H is an isomorphism, then we can use it to prove the isomorphism between G/N and H. This is because an isomorphism between two groups implies that the groups have the same structure and properties, and therefore can be considered equivalent.

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