Monthly Archives: January 2009

Question on automorphism groups

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.

Question on an infinite series

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.)