Representations of group Z_2

  • #1
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Every group needs to have that every element appear only once at each row and each column. But in the case of unfaithful representations of ##Z_2## sometimes we have ##D(e)=1##, ##D(g)=1##. When we write the Caley table we will have that one appears twice in both rows and in both columns. How is that group?
 
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  • #2
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A group representation is an homomorphism ##D\, : \,G\longrightarrow GL(S)##. Both are groups, but they neither need to be isomorphic, nor epimorphic, nor monomorphism. They Cayley tables of these groups have nothing to do with ##D##.
 
  • #3
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This is my question. If I see this table is it a group and why?
[tex]

\begin{array}{l|*{5}{l}}
& 1 & 1 \\
\hline
1 & 1 & 1 \\
1 & 1 & 1 \\

\end{array}

[/tex]
 
  • #4
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This is my question. If I see this table is it a group and why?
[tex]

\begin{array}{l|*{5}{l}}
& 1 & 1 \\
\hline
1 & 1 & 1 \\
1 & 1 & 1 \\

\end{array}

[/tex]
This makes not much sense. ##\{1,1\}=\{1\}##. Do you have one one or two ones?
 
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  • #6
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So I have two elements that are identical.
 
  • #7
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So I have two elements that are identical.
Read this again! ##\{1\}## is a group. If you like, you can build ##\{1\}\times \{1\}##, but I am almost certain that this is not what you meant.
 
  • #8
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Well, this is a representation of ##Z_2## group. So it needs to be a group. If I write just Caley table it is confusing. Because of that, I asked the question. Your answers till now were not helpful.
 
  • #9
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What I meant is that if you have ##Z_2## group with elements ##e## and ##g## and you define
##f(e)=1##,##f(g)=1## could that you say that you have ##Z_2## group with elements ##1## and ##1## and this is unfaithful representation of ##Z_2## group.
 
  • #10
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Well, this is a representation of ##Z_2## group. So it needs to be a group. If I write just Caley table it is confusing. Because of that, I asked the question. Your answers till now were not helpful.
There is no such thing as a Cayley table for a representation.

The Cayley table for ##\mathbb{Z}_2## is
\begin{array}{l|*{5}{l}}
+ & 0 & 1 \\
\hline
0 & 0 & 1 \\
\hline
1 & 1 & 0 \\
\end{array}

As I said earlier, a representation is a group homomorphism ##D\, : \,\mathbb{Z}_2\longrightarrow \operatorname{Aut}(S)## with a set ##S##. Let's assume for simplicity that ##S## is finite, say ##S=\{s_1,\ldots,s_n\}.## Then ##D(0)=\operatorname{id}_S=1## so ##D(0)(s_k)=s_k##. ##D(1)=\sigma## is then a permutation of ##S## and since ##\sigma^2=\sigma\cdot\sigma=D(1)\cdot D(1)=D(1+1)=D(0)=1## it is of order two. The only table I see here is to write
\begin{align*}
D\, : \,\mathbb{Z}_2 &\longrightarrow \operatorname{S}_n\\
0&\longmapsto \operatorname{id}_S = (s_k\longmapsto s_k)\\
1&\longmapsto \sigma=(s_k\longmapsto \sigma(s_k))
\end{align*}

I have the impression that you identified ##0\in \mathbb{Z}_2## with ##\operatorname{id}_S## and ##1\in \mathbb{Z}_2## with ##\sigma\in \operatorname{S}_n## but why? They are in different groups and as such different elements. Since the kernel of ##D## is a subgroup of ##\mathbb{Z}_2##, we don't have many choices. Either this kernel is ##\{0\}## in which case we have a faithful representation and ##\operatorname{Aut}(S)=S_n \trianglerighteq D(\mathbb{Z}_2)\cong \mathbb{Z}_2,## or we have the trivial representation ##D(0)=D(1)=\operatorname{id}_S##.
 
  • #11
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The only difference between them is that they have different elements that correspond to them in the original group. If for instance, we have a set of matrices that obey the rule of some group. Let's say ##S_3##, why I can not write down the multiplication table of those matrices? And what if two of those matrices are completely the same? Could I write a multiplication table for them? If not, why not?
 
  • #12
Infrared
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What I meant is that if you have ##Z_2## group with elements ##e## and ##g## and you define
##f(e)=1##,##f(g)=1## could that you say that you have ##Z_2## group with elements ##1## and ##1## and this is unfaithful representation of ##Z_2## group.
A representation is not a group- it is a (special type of) homomorphism. Given a homomorphism ##f:G\to H##, there is certainly nothing stopping you from considering all products ##f(g)f(g')=f(gg')## for ##g,g'\in G## and writing down a multiplication table, but as noted above this won't be a multiplication table for a group when ##f## isn't injective. If you throw out duplicates, then it will be since the image of a homomorphism is a subgroup of its codomain.
 
  • #13
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This is my question. If I see this table is it a group and why?
[tex]

\begin{array}{l|*{5}{l}}
& 1 & 1 \\
\hline
1 & 1 & 1 \\
1 & 1 & 1 \\

\end{array}

[/tex]
The elements of ##Z_2## are 0 and 1. If the group operation is multiplication, the table above will look like this:
$$ \begin{array}{l|*{5}{l}}
* & 0 & 1 \\
\hline
0 & 0 & 0 \\
1 & 0 & 1 \\

\end{array}$$
 
  • #14
Office_Shredder
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I think the point is that you can take a group representation of ##\mathbb{Z}_2## which maps both 0 and 1 to 1. But then the new group has only one element, not two.
 
  • #15
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But then it's the trivial group, not Z2. (I suppose it could be called C1 or Z1, but I have never heard it called anything but "trivial")
 
  • #16
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I think the OP is very confused about the concept of a representation, and as far as I am concerned, refuses to even think about the correct phrasing. All began with the identification of ##1\in G## with ##1\in \operatorname{Aut}(S)##, and to speak of Cayley tables without distinguishing between ##G,\varphi(G),\operatorname{Aut}(S)##, and in case ##S## is a group, too, then ##S## adds to the list.

You cannot answer an ill-phrased question when simultaneously the questioner insists on the phrasing.
 
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  • #17
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But then it's the trivial group, not Z2. (I suppose it could be called C1 or Z1, but I have never heard it called anything but "trivial")

A representation of a group is just a group homomorphism to the general linear group. Nothing requires it to be injective, that is in fact what a faithful representation is.
 

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