# Representations of group Z_2

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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fresh_42
Mentor
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##.

This is my question. If I see this table is it a group and why?
$$\begin{array}{l|*{5}{l}} & 1 & 1 \\ \hline 1 & 1 & 1 \\ 1 & 1 & 1 \\ \end{array}$$

fresh_42
Mentor
This is my question. If I see this table is it a group and why?
$$\begin{array}{l|*{5}{l}} & 1 & 1 \\ \hline 1 & 1 & 1 \\ 1 & 1 & 1 \\ \end{array}$$
This makes not much sense. ##\{1,1\}=\{1\}##. Do you have one one or two ones?

I have ##(\{1,1\},\cdot)##.

So I have two elements that are identical.

fresh_42
Mentor
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.

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.

weirdoguy
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.

fresh_42
Mentor
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##.

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?

Infrared
Gold Member
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.

Mark44
Mentor
This is my question. If I see this table is it a group and why?
$$\begin{array}{l|*{5}{l}} & 1 & 1 \\ \hline 1 & 1 & 1 \\ 1 & 1 & 1 \\ \end{array}$$
The elements of ##Z_2## are 0 and 1. If the group operation is multiplication, the table above will look like this:
$$\begin{array}{l|*{5}{l}} * & 0 & 1 \\ \hline 0 & 0 & 0 \\ 1 & 0 & 1 \\ \end{array}$$

Office_Shredder
Staff Emeritus
Gold Member
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.

Staff Emeritus
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")

fresh_42
Mentor
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.