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Math Help - Abstract-several questions

  1. #1
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    Abstract-several questions

    Well, I'm trying to solve some old exams in abstract algebra, and I need your help in the following:

    Question 1:
    Let  U(Z/15Z) be the multiplicative group of the invertible elemenets in  Z/15Z . Write  U(Z/15Z) as a product of cyclic groups.

    My try:
    It's easy to show that x is in U iff gcd(x,15)=1... Hence:
    U={1,2,8,4,7,13,11,14} ... If we'll take the element 2 to be a generator we'll get the cyclic group {1,2,4,8} ... If we'll take <7> we'll get {1,7,4,13} etc... I can't figure out how to write U as a product of cyclic groups when the intersection of each two cyclic subgroups of U is "bigger" than {1}...


    Question 2:
    Prove that  Z/nZ has only one maximal ideal (which isn't trivial) iff n is a power of a prime number.

    Question 3:
    Find a 3-sylow subgroup of  S_{8} and find a group that is isomorphic to it.

    My try:
    We know  o(S_{8}) = 8! . Hence, a 3-sylow subgroup H is from order  o(H)= 3^{2}=9 . If we'll be able to find an element in  S_{8} from that order- we're done...But is there any element of that order? How should I solve this one?


    Thanks a lot!
    Last edited by WannaBe; February 9th 2010 at 06:11 AM.
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  2. #2
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    Quote Originally Posted by WannaBe View Post
    Well, I'm trying to solve some old exams in abstract algebra, and I need your help in the following:

    Question 1:
    Let  U(Z/15Z) be the multiplicative group of the invertible elemenets in  Z/15Z . Write  U(Z/15Z) as a product of cyclic groups.

    My try:
    It's easy to show that x is in U iff gcd(x,15)=1... Hence:
    U={1,2,8,4,7,13,11,14} ... If we'll take the element 2 to be a generator we'll get the cyclic group {1,2,4,8} ... If we'll take <7> we'll get {1,7,4,13} etc... I can't figure out how to write U as a product of cyclic groups when the intersection of each two cyclic subgroups of U is "bigger" than {1}...
    If \mathbb{Z}/m\mathbb{Z} \cong \mathbb{Z}/m_1\mathbb{Z} \oplus \mathbb{Z}/m_2\mathbb{Z} \oplus \cdots \oplus \mathbb{Z}/m_i\mathbb{Z} , then U(\mathbb{Z}/m\mathbb{Z}) \cong U(\mathbb{Z}/m_1\mathbb{Z}) \times U(\mathbb{Z}/m_2\mathbb{Z}) \times \cdots \times U(\mathbb{Z}/m_i\mathbb{Z}) .

    We see that if each m_k is a prime number, then each U(\mathbb{Z}/m_k\mathbb{Z}) is a cyclic group.

    Thus, U(\mathbb{Z}/15\mathbb{Z}) \cong U(\mathbb{Z}/3\mathbb{Z}) \times U(\mathbb{Z}/5\mathbb{Z}).
    Question 2:
    Prove that  Z/nZ has only one maximal ideal (which isn't trivial) iff n is a power of a prime number.
    If n is a power of prime number, then \mathbb{Z}/n = \mathbb{Z}/p^i\mathbb{Z}. The maximal ideal of \mathbb{Z}/p^i\mathbb{Z} is \mathbb{Z}/p^{i-1}\mathbb{Z} for i>=2, which is unique for n=p^i.
    Question 3:
    Find a 3-sylow subgroup of  S_{8} and find a group that is isomorphic to it.

    My try:
    We know  o(S_{8}) = 8! . Hence, a 3-sylow subgroup H is from order  o(H)= 3^{2}=9 . If we'll be able to find an element in  S_{8} from that order- we're done...But is there any element of that order? How should I solve this one?


    Thanks a lot!
    H={e, (1,2,3), (1,3,2), (4,5,6), (4,6,5), (1,2,3)(4,5,6), (1,2,3)(4,6,5), (1,3,2)(4,5,6), (1,3,2)(4,6,5)}. Other subgroups of order 9 in S_8 can be found similarly.
    Last edited by aliceinwonderland; February 10th 2010 at 02:15 AM. Reason: Correction
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  3. #3
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    Thanks a lot man!
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  4. #4
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    Quote Originally Posted by WannaBe View Post
    Thanks a lot man!
    Please check a modification of Q3.
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  5. #5
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    Quote Originally Posted by aliceinwonderland View Post
    Please check a modification of Q3.
    modification of Q3?
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  6. #6
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    Quote Originally Posted by WannaBe View Post
    modification of Q3?
    Yeah, I had edited my answer for Q3. Even though 9 | 8! , I was not able to find an order 9 subgroup in S_8.

    Sorry for any confusion. My English is no good.
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  7. #7
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    Oh , you meant Question 3 LOL ... I wasn't sure about it.... Yep, I saw your "modification"... But we know that there are 2 groups of order 9 up to isomorphism: C_3 x C_3 and C_9... We don't have an element of order 9 in S8 hence the only subgroup of order 9 in S8 is C_3xC_3 ... (I think so...)

    Thanks a lot anyway !
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