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I've got this homework question for Group Theory and it states: $A=\begin{bmatrix} 1 \ 2 \ 3 \ 4 \ 5 \\ 1 \ 4 \ 3 \ 2 \ 5 \\ \end{bmatrix}$ and $B=\begin{bmatrix} 1 \ 2 \ 3 \ 4 \ 5 \\ 3 \ 5 \ 1 \ 2 \ 4 \\ \end{bmatrix}$ and asks me to calculate $AB$ as well as the inverse of $A$.

The trouble I'm having is that so far with my limited knowledge I thought that the columns in a had to equal the rows in b, as well as only a square matrix had an inverse. The second part of the question then asks to solve ax=b. Can somebody help explain what I'm missing? I don't believe its a mistake as a similar question appears in last years problem set. Maybe I should interpret a and b as sets with two rows or something? All help is much appreciated!

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3 Answers

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These appear to be elements of the symmetric group, and not matrices. (1 2 3 4 5; 1 4 3 2 5) is the permutation that sends 1 to 1, 2 to 4, 3 to 3, 4 to 2, and 5 to 5. Let me know if you need more help.

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I don't think you're meant to view these as matrices. I think the notation $(1,2,3,4,5;1,4,3,2,5)$ means that $1\to 1$, $2\to 4$, $3\to 3$, $4\to 2$ and $5\to 5$. You're meant to compose the permutations.

For example, if $a = (1,2,3;1,3,2)$ and $b = (1,2,3;3,1,2)$, then under $b$, $1\to 3$, $2\to 1$ and $3\to 2$. Under $a$, $1\to 1$, $2\to 3$ and $3\to 2$. Multiplying here means: do the permutation $b$ then do the permutation $a$. The reason that we do it left to right is that it is compositions of permutations, just like compositions of functions. $(f\circ g)(x) = f(g(x))$, meaning first you do $g(x)$, then you apply $f$ to that. Same logic applies here.

So in this example, $1\stackrel{b}{\to} 3 \stackrel{a}{\to} 2$, $2\stackrel{b}{\to} 1 \stackrel{a}{\to} 1$ and $3\stackrel{b}{\to} 2\stackrel{a}{\to} 3$. Thus $ab = (1,2,3;2,1,3)$. Can you see how to proceed for your problem?

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The matrix notation for permutations: $ A=\begin{pmatrix}1&2&3&4&5\\1&4&3&2&5 \end{pmatrix} $ consists in writing in the first line the set to be permuted (here natural numbers from $1$ to $5$ and in the second line the images of the elements in the first line. Conventionally, the first line is written in increasing order.

So here, when you compute the image of $1$ by $AB$, you first begin with sending it to $3$ by $B$, then sending $3$ to itself by $A$; so $1$ goes to $3$ by $AB$.

As to the inverse of $A$, it is the reverse transformation, so you should exchange the first and second lines in $A$, except the first line will not be written in increasing order, so you'll have to rewrite the inverse of $A$.

Finally $AX=B\iff X=A^{-1}B$. Beware the product of two permutations is not commutative in general.

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