9367 #51 #55 #56

Forum for the GRE subject test in mathematics.
breezeintopl
Posts: 16
Joined: Sun Jul 19, 2009 12:29 pm

9367 #51 #55 #56

Postby breezeintopl » Thu Nov 05, 2009 9:10 pm

#51
why there is only one automorphism? I mean...is \phi(x)=-x a second one?
what is the proof of this problem? the same to the <cracking P250 ex6.22>?
#55
I think I have seen the answer before but I just cannot find it><
#56
omg, I've got no idea about this Qs...what is S'?and...

thanks a lot~~

mtey
Posts: 22
Joined: Tue Oct 27, 2009 12:30 pm

Re: 9367 #51 #55 #56

Postby mtey » Fri Nov 06, 2009 5:43 am

#51

There is a significant difference between #51 and Leduc's example, namely in #51 they want \phi to be one-to-one and onto. This implies that \phi is not trivial. After that like in Leduc's example we can show \phi(1) = 1, \phi(n) = n, \phi(\frac{p}{q}) = \phi(p)\phi(\frac{1}{q}) = \frac{\phi(p)}{\phi(q)} = \frac{p}{q}

PS \phi(x) = -x is not a homomorphism as 1 = \phi(-1) = \phi(-1*1) = \phi(-1)\phi(1) = -1

#55

if gcd(a,b) = 1 then there exists k, m \in \mathbb{Z} such that ka + mb =1 Thus we can easily eliminate (A), (D) (gcd(p^q, q^p) = 1), and (C), (B) (gcd(p+q, pq) = 1) since they do not generate proper subgroups of \mathbb{Z}. (E) is obviously fine for p\mathbb{Z}.

#56

This is a tricky question.

I. Is true since for all x \in A such that every open set that contains them contains at least one other point in A, obviously have the same property for (A \cup B), thus A' \cup B' \subset (A \cup B)'.
Conversely if x \in A \cup B holds the property that every open set that contains it contains at least one other point from A \cup B, we can assume WLOG that all of those open sets hold a point from A, but this means that x \in A', so (A \cup B)' \subset A' \cup B'

II. A nice counterexample is A = \mathbb{Q} \cap [0,1], B = [0,1] - A

III. Is true since \operatorname{cl}(A) = A \cup A' = A, which means that A is closed.

IV. A = \emptyset is counterexample.
Last edited by mtey on Sat Nov 14, 2009 8:32 am, edited 1 time in total.

breezeintopl
Posts: 16
Joined: Sun Jul 19, 2009 12:29 pm

Re: 9367 #51 #55 #56

Postby breezeintopl » Fri Nov 06, 2009 7:13 am

mtey wrote:#51

There is a significant difference between #51 and Leduc's example, namely in #51 they want \phi to be one-to-one and onto. This implies that \phi is not trivial. After that like in Leduc's example we can show \phi(1) = 1, \phi(n) = n, \phi(\frac{p}{q}) = \phi(p)\phi(\frac{1}{q}) = \frac{\phi(p)}{\phi(q)} = \frac{p}{q}

PS \phi(x) = -x is not a homomorphism as 1 = \phi(-1) = \phi(-1*1) = \phi(-1)\phi(1) = -1

#55

if gcd(a,b) = 1 then there exists k, m \in \mathbb{Z} such that ka + mb =1 Thus we can easily eliminate (A), (D) (gcd(p^q, q^p) = 1), and (C), (B) (gcd(p+q, pq) = 1) since they are not even proper subgroups of \mathbb{Z}. (E) is obviously fine for p\mathbb{Z}.

#56

This is a tricky question.

I. Is true since for all x \in A such that every open set that contains them contains at least one other point in A, obviously have the same property for (A \cup B), thus A' \cup B' \subset (A \cup B)'.
Conversely if x \in A \cup B holds the property that every open set that contains it contains at least one other point from A \cup B, we can assume WLOG that all of those open sets hold a point from A, but this means that x \in A', so (A \cup B)' \subset A' \cup B'

II. A nice counterexample is A = \mathbb{Q} \cap [0,1], B = [0,1] - A

III. Is true since \operatorname{cl}(A) = A \cup A' = A, which means that A is open.

IV. A = \emptyset is counterexample.


wow,cool~~~you are so nice~~~

ps:when I learn topology, S' is defined as the collection of all the limit points of S.
but, emm...I don't know whether the two definition are the same.

jayre
Posts: 11
Joined: Wed Nov 04, 2009 10:30 pm

Re: 9367 #51 #55 #56

Postby jayre » Fri Nov 06, 2009 2:10 pm

breezeintopl wrote:ps:when I learn topology, S' is defined as the collection of all the limit points of S.
but, emm...I don't know whether the two definition are the same.

That sounds right to me.

yoyobarn
Posts: 80
Joined: Sun Dec 19, 2010 7:01 am

Re: 9367 #51 #55 #56

Postby yoyobarn » Mon Dec 19, 2011 12:11 pm

mtey wrote:#51

There is a significant difference between #51 and Leduc's example, namely in #51 they want \phi to be one-to-one and onto. This implies that \phi is not trivial. After that like in Leduc's example we can show \phi(1) = 1, \phi(n) = n, \phi(\frac{p}{q}) = \phi(p)\phi(\frac{1}{q}) = \frac{\phi(p)}{\phi(q)} = \frac{p}{q}

PS \phi(x) = -x is not a homomorphism as 1 = \phi(-1) = \phi(-1*1) = \phi(-1)\phi(1) = -1



Can anyone explain what is the one distinct automorphism?

Also, how do we get \phi (\frac{1}{q})=\frac{1}{\phi (q)}?

Thanks a lot.

Ellen
Posts: 3
Joined: Sat Aug 10, 2013 3:56 pm

Re: 9367 #51 #55 #56

Postby Ellen » Sat Aug 10, 2013 4:07 pm

cgi-bin/mathtex.cgi?\phi%20(\frac{1}{q})=\frac{1}{\phi%20(q)}

Since every element in a field has a multiplicative inverse by definition, and f(e) = e, f(1/q*q) = f(1/q)*f(q) = e. Thus f(1/q) = e/f(q).

Sorry I don't (yet) have latex. :( Trust me, I can't wait...




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