Understanding Unfaithful Representations of Z_2 in the Caley Table

[LagrangeEuler]

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?

 

[fresh_42]

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 $D$.

 

[LagrangeEuler]

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]

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?

 

[LagrangeEuler]

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

 

[LagrangeEuler]

So I have two elements that are identical.

 

[fresh_42]

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.

 

[LagrangeEuler]

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.

 

[LagrangeEuler]

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]

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 $D(0)=D(1)=\operatorname{id}_S$.

 

[LagrangeEuler]

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]

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]

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]

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.

 

[Vanadium 50]

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]

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.

 

[Office_Shredder]

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.