Abstract-several questions

• Feb 9th 2010, 01:33 AM
WannaBe
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!
• Feb 9th 2010, 11:42 PM
aliceinwonderland
Quote:

Originally Posted by WannaBe
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, $U(\mathbb{Z}/15\mathbb{Z}) \cong U(\mathbb{Z}/3\mathbb{Z}) \times U(\mathbb{Z}/5\mathbb{Z})$.
Quote:

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 $n=p^i$.
Quote:

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.
• Feb 10th 2010, 12:39 AM
WannaBe
Thanks a lot man!
• Feb 10th 2010, 12:47 AM
aliceinwonderland
Quote:

Originally Posted by WannaBe
Thanks a lot man!

Please check a modification of Q3.
• Feb 10th 2010, 01:10 AM
WannaBe
Quote:

Originally Posted by aliceinwonderland
Please check a modification of Q3.

modification of Q3?
• Feb 10th 2010, 01:27 AM
aliceinwonderland
Quote:

Originally Posted by WannaBe
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.
• Feb 10th 2010, 01:53 AM
WannaBe
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 !