Results 1 to 4 of 4

Thread: Cyclic Groups

  1. #1
    Member
    Joined
    Dec 2009
    Posts
    225

    Cyclic Groups

    The question has 3 parts:


    My Attempt:

    (a) Since the cyclic group generated by a is $\displaystyle \left\langle a \right\rangle = \{ za | z \in \mathbb{Z} \}$, I think it would follow that for distinct $\displaystyle a_1,a_2,...,a_n \in \mathbb{Z}$

    $\displaystyle \left\langle a_1,...,a_n \right\rangle = \{ z_1a_1 +...+z_na_n\}$

    but I don't know how to actually prove this. I appreciate any hints to get started.

    (b) Though I'm not sure if it helps... but I think in general $\displaystyle \left\langle ma \right\rangle \cap \left\langle na \right\rangle = \left\langle ka \right\rangle$ where $\displaystyle k =lcm(m,n) \mod\ \mathbb{Z}$. Is that correct?

    (c) Is the gcd(a,b) required to be 1 (so that a and b are relatively prime)? If so I get

    $\displaystyle \left\langle a,b \right\rangle = \{ au+bv | u,v \in \mathbb{Z}\}= \left\langle h \right\rangle$

    $\displaystyle \implies au+bv=1$

    Is "h" equal to 1?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Swlabr's Avatar
    Joined
    May 2009
    Posts
    1,176
    Part (a) is one of those annoying questions which is quite genuinely obvious and so annoying to prove. I would try inducting on 'n', the number of generators. If that doesn't work, I wouldn't worry too much about it!

    Yes, in (b) it is the least common multiple you are after.

    For (c) you are on the right lines, but generalise your result of gcd=1 to arbitrary integers.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Dec 2009
    Posts
    225
    Quote Originally Posted by Swlabr View Post
    Part (a) is one of those annoying questions which is quite genuinely obvious and so annoying to prove. I would try inducting on 'n', the number of generators. If that doesn't work, I wouldn't worry too much about it!
    I plan to prove it in 3 steps but I'm having some trouble:

    1. $\displaystyle \forall z_1,...,z_n$, we have $\displaystyle z_1a_1+z_2a_2+...+z_na_n \in \left\langle a_1,...,a_n \right\rangle$

    Therefore $\displaystyle \imples \{ z_1a_1+z_2a_2+...+z_na_n | z \in \mathbb{Z} \} \subseteq \left\langle a_1,...,a_n \right\rangle$

    The problem is that I don't see how to prove that every linear combination $\displaystyle z_1a_1+z_2a_2+...+z_na_n$ must be in $\displaystyle \left\langle a_1,...,a_n \right\rangle$.

    2. $\displaystyle \{ z_1a_1+z_2a_2+...+z_na_n | z \in \mathbb{Z} \}$ is a subgroup of $\displaystyle \mathbb{Z}$ containing $\displaystyle \left\langle a_1,...,a_n \right\rangle$.

    To prove that it's a subgroup is it possible to use the "one step test"?

    3. If I can then show that $\displaystyle \left\langle a_1,...,a_n \right\rangle$ is the smallest subgroup of $\displaystyle \mathbb{Z}$ containing $\displaystyle a_1,...,a_n $ with $\displaystyle \{ z_1a_1+z_2a_2+...+z_na_n | z \in \mathbb{Z} \} \supseteq \left\langle a_1,...,a_n \right\rangle$

    I have shown that

    $\displaystyle \{ z_1a_1+z_2a_2+...+z_na_n | z \in \mathbb{Z} \} \subseteq \left\langle a_1,...,a_n \right\rangle$
    $\displaystyle \{ z_1a_1+z_2a_2+...+z_na_n | z \in \mathbb{Z} \} \supseteq \left\langle a_1,...,a_n \right\rangle$

    $\displaystyle \therefore \{ z_1a_1+z_2a_2+...+z_na_n | z \in \mathbb{Z} \} = \left\langle a_1,...,a_n \right\rangle$

    Yes, in (b) it is the least common multiple you are after.
    Actually I meant:

    $\displaystyle \left\langle a \right\rangle \cap \left\langle b \right\rangle = \left\langle k \right\rangle$

    where k = lcm(a,b) mod Z = <g>.

    So I think here I have to go both ways i.e: show that

    $\displaystyle \left\langle a \right\rangle \cap \left\langle b \right\rangle \leq lcm(a,b)$ .....(1)

    $\displaystyle \left\langle a \right\rangle \cap \left\langle b \right\rangle \geq lcm(a,b)$ .....(2)

    Could you show me how to go about proving one of them? I absolutely don't know where to start...

    For (c) you are on the right lines, but generalise your result of gcd=1 to arbitrary integers
    How? It follows from part (a) that

    $\displaystyle \left\langle a,b \right\rangle = \{ z_aa+z_bb | z_a,z_b \in \mathbb{Z}\}\in gcd (a,b)$

    $\displaystyle \exists z_a, z_b \in \mathbb{Z}$ with $\displaystyle z_a a + z_bb=gcd(a,b)$

    I think similar to part (b) I have to prove in both directions:

    $\displaystyle \left\langle a,b \right\rangle \leq \left\langle gcd(a,b) \right\rangle$
    $\displaystyle \left\langle a,b \right\rangle \geq \left\langle gcd(a,b) \right\rangle$

    We can let $\displaystyle h=gcd(a,b)$ then we can write a=ch and b=dh where c,d are integers.

    If $\displaystyle x \in \left\langle a \right\rangle \implies x \in \left\langle ch \right\rangle $

    Also if $\displaystyle x \in \left\langle b \right\rangle \implies x \in \left\langle dh \right\rangle $

    Then what can we say next?

    P.S. By the "$\displaystyle \leq$" notation I mean 'a subgroup of', not 'less than or equal to'.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Dec 2009
    Posts
    225
    But if my method is hopeless, then you could show me how to do it with induction.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Automorphism groups of cyclic groups
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: Aug 15th 2011, 09:46 AM
  2. Cyclic groups and gcd
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Mar 12th 2011, 08:32 AM
  3. Cyclic Groups
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: Nov 3rd 2009, 07:26 AM
  4. cyclic groups
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: Oct 2nd 2008, 05:14 AM
  5. Cyclic groups?
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: Dec 27th 2007, 12:05 PM

Search Tags


/mathhelpforum @mathhelpforum