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I was having some trouble understanding the following question:

Q: Let $G=\{1,-1\}$ endowed with the classical multiplication of integers.

Describe the left multiplication $\ell_{(1,-1)}$ by the element $(1,-1)$ in $G\times G$ as a permutation of the element of the set $G\times G$.

So I have been researching and I understand I have to write it as the product of disjoint cycles somehow but I have no idea how to apply this theorem to actual examples. If anyone could perhaps explain it in a simple way I would be very grateful.

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In a group the function $l_g:x\mapsto gx$ is a permutation of the set of group elements.

The question asks you to describe this permutation in the particular case where the group is the direct product of the group $G$ described with itself and the particular $g$ in the above expression is the element (1,-1) in that group.

You can describe the permutation by showing the result of the permutation on each element of $G\times G$ (there are only four).

E.g. $l_{(1,-1)}:(1,1)\mapsto (1,-1)\ \ (1,-1)\mapsto (1,1)\ldots$ (complete the description by replacing the dots by the mappings for the remaining elements.)

Or as a cycle decomposition of the permutation:

$l_{(1,-1)}=((1,1),(1,-1))(\ldots)$ (again complete by replacing the dots with the content of the other cycle.)

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    Sorry if this is a trivial question but I don't understand the $(1,-1)$ notation, how are there four elements? Doesn't the group only consist of $-1$ and $1$ ? – 2017-02-13
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    If I'm understanding correctly, is it just listing each permutation, so would the remaining cycles be $(-1,-1)$ and $(-1,1)$ ? – 2017-02-13
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    For two groups $G$ and $H$ the definition of $G\times H$ is the set of ordered pairs where the first element is taken from $G$ and the second from $H$, together with a specific group operation on that set. The individual ordered pairs are written as $(g,h)$ where $g\in G$ and $h\in H$. So the notation (1,-1) means the ordered pair with first element 1 and second element -1. (In this case both are from the group $G$ you give, because this is both the fis and second group in the direct product. – 2017-02-13
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    The above comment was truncated - the rest follows: – 2017-02-13
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    The group $G$ has only two elements but there are four in the set of ordered pairs, viz. (1,1), (1,-1), (-1,1) and (-1,-1). – 2017-02-13
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    The answer to your second query is no. A cycle is written $(a,b,c,\ldots ,n)$ and means the permutation which maps $a$ to $b$, $b$ to $c$ and so on until you reach $n$ which is mapped back to $a$. – 2017-02-13
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    continued ... If (-1,1) were to be understood as a cycle (the notation for a cycle with two elements is identical with the notation for an ordered pair, so it can be taken two ways), the the elements permuted would be -1 and 1 which are elements of $G$ *not* $G\times G$. The permutation required is a permutation of $G\times G$. In the cycle decomposition I sketched in the answer the outer parentheses denote a cycle, while the inner ones denote ordered pairs. – 2017-02-13
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    I hope I'm following now but would the second cycle then be: $((-1,1),(-1,-1))$ ? – 2017-02-13
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    Yes, exactly. The full cycle decomposition would then be written: ((1,1),(1,-1))((-1,1),(-1,-1)) – 2017-02-13
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    You appear to have resolved your problem. Is the answer acceptable?I notice you didn't accept it. – 2017-02-14
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    Ah yes my mistake. Thank you! – 2017-02-14