2009 January « Bloomsbury Journal

Hello Prof Kim

I got a question about groups and ring ( algebra 4). Can you explain to me what it means to “describe $Aut(C_n)$ explicitly” for n = 1, 2,3, ….?

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Reply:

In many parts of mathematics, one uses the notation $Aut(S)$ to denote the automorphisms of $S$ , where $S$ is a set usually with some extra structure. Thus, $Aut(S)$ consists of maps

$f:S \rightarrow S$

with the property that

(1) $f$ has an inverse;

(2) $f$ is compatible with whatever structure $S$ has, if any.

So when you’re being asked to describe, say, $Aut(C_3)$ explicitly, you are being asked to describe all invertible maps

$f: \{0,1,2 \} \rightarrow \{0,1,2\}$

that are compatible with the group structure. One point you shouldn’t get confused by is that $Aut(S)$ isl *always* a group under composition, whatever structure $S$ has. If $S$ is itself a group, $Aut(S)$ will in general be some other group.

Note that if we had left out the group structure and just considered the set $\{0,1,2\}$ , then the automorphisms would be isomorphic to (‘essentially the same as’) $S_3$ , the symmetric group on three letters. But with the group structure, you need to be more careful. Not all permutations will preserve the group structure. Think about it a bit and ask again if it’s still confusing.

Hello Sir,

I was in your algebra 3 course last year and found this blog useful so I was hoping you could provide me with some assistance on the following problem from my Theory of Numbers Course.

How would you show that Sigma(1/p^2) is less that or equal to 1. Where p is a prime.

I would really appreciate any help you could give me.

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Reply:

First of all, it’s better to say that the $p$ in the sum *runs over* the set of primes. If you say $p$ is a prime, it sounds like we’re speaking just of one.

Anyways, I’m hoping you learned a bit about the Riemann zeta function

$\zeta (s)=\sum_{n=1}^{\infty} n^{-s}.$

It is easy to see that this sum converges for $Re(s)>1$ and, importantly, can be written also as an infinite product in this range:

$\zeta(s)=\prod_p\frac{1}{(1-p^{-s})},$

where again the $p$ runs over the primes. In particular,

$\zeta (2)=\sum_{n=1}^{\infty} n^{-2}=\prod_p\frac{1}{(1-p^{-2})}$

If you write the last quantity as

$\prod_p(1+p^{-2}+p^{-4}+\cdots),$

and expand the product, you will see that it’s greater than

$1+\sum_p p^{-2}.$

Thus, the sum you’re interested in has shown up. Hence,

$\sum_p p^{-2} < \zeta(2) -1.$

Actually, it’s possible to evaluate $\zeta(2)$ precisely, and get $\pi^2/6$ . However, for your inequality, it’s not necessary. All you need to know is $\zeta(2) \leq 2.$ Try to show this by bounding the sum for $\zeta(2)$ by an integral. (Recall the idea in the integral test for convergence of a positive series.)

Bloomsbury Journal

January 2009

Question on automorphism groups

Question on an infinite series

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