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Problem: Give an example of a group G, subgroup H, and an element
a in G such that aHa^(-1) is a subset of H but aHa^(-1) is
NOT equal to H.

[Problem #8 p53 in Topics in Algebra, by Herstein (Second Ed, 1975)]

Any hint or suggestion would be most welcome.

Thank you,

--Bhatia


there exists a prime between x^2 and (x +1)^2.
...
assume... (x^2)/ln(x^2) +1 < ((x +1)^2)/ln((x +1)^2) using the Prime Number Theorem,
...
(x^2 +ln(x^2))/ln(x^2) < (x^2 +2x +1)/ln((x +1)^2) combining the LHS with the +1,
...
x^2 +2*ln(x) < (x^2 +2x +1)*[(ln(x^2))/(ln((x +1)^2)] multiplying both sides by ln(x^2),
...
but the limit [(ln(x^2))/(ln((x +1)^2)] -> 1 almost immediately, and we have 2*ln(x) < 2x +1, and taking the derivative wrt 'x', we have... (d/dx)[2*ln(x)] < (d/dx)[2x +1], or... 2 < 2x; so our beginning statement has to be true for x >= 2, and it is...
...
x= 2; LHS= 3.885 < 4.096 = RHS, and x= 3; LHS= 5.096 < 5.770 = RHS...and so on, and so on. we simply check between 1 and 4 to see that the primes 2 and 3 exist!
...
nice! right? Bill
www.oddperfectnumbers.com

Please post your incorrect proofs in your own threads (or not at all). Don't hijack other people's unrelated threads.

Thank you, Yark.

I am afraid I do not understand what you mean.

Please note that the Herstein (2nd Ed) Problem #8 p53
is before the homomorphisms are defined. Uptil now,
we have only seen the definition of a normal subgroup
and a quotient group.

Any way to think about the problem without refering to
automorphisms?

I do appreciate your help.

-- bhatia

bhatia writes:

>I am afraid I do not understand what you mean.

The automorphism I mentioned can be done by conjugation in a suitable extension, and will have the required property.

>Please note that the Herstein (2nd Ed) Problem #8 p53
>is before the homomorphisms are defined. Uptil now,
>we have only seen the definition of a normal subgroup
>and a quotient group.

Ah, I didn't realise that.

>Any way to think about the problem without refering to
>automorphisms?

The example I was thinking of can be done without referring to automorphisms. Let G be the group of all functions Q -> Q of the
form ax+b, where a and b are rational, and a is non-zero. Let H be the subgroup consisting of integer translations (that is, functions of the form x+n, where n is an integer).

Let F(x,y) be a free group on two generators. See

http://en.wikipedia.org/wiki/Free_group

if you don't know what a free group is. For any natural n consider a subgroup

H(n) = < x^{m}yx^{-m} | m>=n >

It can be easily seen that the following rules hold:

1) xH(n)x^{-1} = H(n+1)
2) H(n+1) is properly contained in H(n)

So we have an example
joking

Here's a hint: there's an automorphism of Q that maps Z to 2Z.

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