#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~~
9367 #51 #55 #56
Re: 9367 #51 #55 #56
#51
There is a significant difference between #51 and Leduc's example, namely in #51 they want $$\phi$$ to be onetoone 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.
There is a significant difference between #51 and Leduc's example, namely in #51 they want $$\phi$$ to be onetoone 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.

 Posts: 16
 Joined: Sun Jul 19, 2009 12:29 pm
Re: 9367 #51 #55 #56
wow,cool~~~you are so nice~~~mtey wrote:#51
There is a significant difference between #51 and Leduc's example, namely in #51 they want $$\phi$$ to be onetoone 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.
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
That sounds right to me.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.
Re: 9367 #51 #55 #56
Can anyone explain what is the one distinct automorphism?mtey wrote:#51
There is a significant difference between #51 and Leduc's example, namely in #51 they want $$\phi$$ to be onetoone 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 $$1 = \phi(1) = \phi(1*1) = \phi(1)\phi(1) = 1$$
Also, how do we get $$\phi (\frac{1}{q})=\frac{1}{\phi (q)}$$?
Thanks a lot.
Re: 9367 #51 #55 #56
http://www.mathematicsgre.com/cgibin/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...
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...