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I am working through Maxfield and Maxfield's Abstract Algebra and Solution by Radicals, and the following exercise is included after discussing permutations and Cayley's Theorem:

Two rows of a certain Cayley table are

r q s 1 u v t w

q s 1 u v t w r

he rows show the eight distinct group members. The group identity is "1." Deduce the rest of the table.

At first, I assumed that the table was written in 1qrstuvw order, but that cannot be true, as it would cause w to occur twice in one column of the Cayley table.

Next, I tried assuming that the table was in abcdefgh order and finding the table by brute force, but I couldn't find out how to make that work beyond identifying inverses.

Finally, I tried to use the isomorphism between the permutation group and the relevant subgroup of $S_8$ used in the proof of Cayley's Theorem, but I couldn't figure out how to get started, as I don't know the order of the elements in the Cayley table.

Thanks so much!!

J. W. Tanner
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RHyp
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  • One of the columns must be $1$. So at least you know there are only $8$ possibilities for which elements the first two rows correspond to. – K. Jiang Jun 16 '24 at 23:59

2 Answers2

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Each group $G$ is isomorphic to a group $H$ of permutations via Cayley's Theorem.

The elements of $H$ are given by permuting those of $G$ according to the rows (or columns) of the Cayley table with respect to left (or right, respectively) multiplication by elements of $G$. See the standard proof of Cayley's Theorem for details.

It follows from the row

$$\hat\sigma=(r,q, s, 1, u, v, t, w)$$

and

$$\hat\tau=(q, s, 1, u, v, t, w, r)$$

that a cyclic shift of the elements has occurred, of order eight.

Thus $G\cong \Bbb Z_8$ because all cyclic groups of a given order are isomorphic.

The table is

$$\begin{array}{c|cccccccc} G & 1 & u & v & t & w & r & q & s \\ \hline 1 & 1 & u & v & t & w & r & q & s \\ u & u & v & t & w & r & q & s & 1 \\ v & v & t & w & r & q & s & 1 & u\\ t & t & w & r & q & s & 1 & u & v\\ w & w & r & q & s & 1 & u & v & t\\ r & \color{purple}r & \color{purple}q & \color{purple}s & \color{purple}1 & \color{purple}u & \color{purple}v & \color{purple}t & \color{purple}w\\ q & \color{orange}q & \color{orange}s & \color{orange}1 & \color{orange}u & \color{orange}v & \color{orange}t & \color{orange}w & \color{orange}r\\ s & s & 1 & u & v & t & w & r & q\\ \end{array}.$$

This was found by cyclically permuting the rows further and starting with the identity row for aesthetic reasons.

Shaun
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  • Your table was really helpful overall, but I'm not quite following why $\sigma^2=\tau$, and how order eight means that $G\cong \mathbb{Z}_8$. – RHyp Jun 17 '24 at 00:52
  • I really shouldn't do mathematics when tired, @RHyp; I messed it up a bit. Here is a new version. I hope it's clearer now! – Shaun Jun 17 '24 at 01:07
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Note that one of the rows is a one character right shift of the other. You should be able to justify that the other six rows are the other cyclic shifts of this order. The one that starts with $1$ is the identity. You can then read off the column headings. The row headings are the first element in the rows you have.

Ross Millikan
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