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9367 #51 #55 #56

Posted: Thu Nov 05, 2009 9:10 pm
by breezeintopl
#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~~

Re: 9367 #51 #55 #56

Posted: Fri Nov 06, 2009 5:43 am
by mtey
#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.

Re: 9367 #51 #55 #56

Posted: Fri Nov 06, 2009 7:13 am
by breezeintopl
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.

Re: 9367 #51 #55 #56

Posted: Fri Nov 06, 2009 2:10 pm
by jayre
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.

Re: 9367 #51 #55 #56

Posted: Mon Dec 19, 2011 12:11 pm
by yoyobarn
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.

Re: 9367 #51 #55 #56

Posted: Sat Aug 10, 2013 4:07 pm
by Ellen
http://www.mathematicsgre.com/cgi-bin/m ... \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...