Thread: 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 $\displaystyle U(Z/15Z) $ be the multiplicative group of the invertible elemenets in $\displaystyle Z/15Z $ . Write $\displaystyle 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 $\displaystyle 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 $\displaystyle S_{8} $ and find a group that is isomorphic to it.

My try:
We know $\displaystyle o(S_{8}) = 8! $ . Hence, a 3-sylow subgroup H is from order $\displaystyle o(H)= 3^{2}=9 $. If we'll be able to find an element in $\displaystyle 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!

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 $\displaystyle U(Z/15Z) $ be the multiplicative group of the invertible elemenets in $\displaystyle Z/15Z $ . Write $\displaystyle 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 $\displaystyle \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 $\displaystyle 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 $\displaystyle m_k$ is a prime number, then each $\displaystyle U(\mathbb{Z}/m_k\mathbb{Z})$ is a cyclic group.

Thus, $\displaystyle U(\mathbb{Z}/15\mathbb{Z}) \cong U(\mathbb{Z}/3\mathbb{Z}) \times U(\mathbb{Z}/5\mathbb{Z})$.

Question 2:
Prove that $\displaystyle 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 $\displaystyle \mathbb{Z}/n = \mathbb{Z}/p^i\mathbb{Z}$. The maximal ideal of $\displaystyle \mathbb{Z}/p^i\mathbb{Z}$ is $\displaystyle \mathbb{Z}/p^{i-1}\mathbb{Z}$ for i>=2, which is unique for $\displaystyle n=p^i$.

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

My try:
We know $\displaystyle o(S_{8}) = 8! $ . Hence, a 3-sylow subgroup H is from order $\displaystyle o(H)= 3^{2}=9 $. If we'll be able to find an element in $\displaystyle 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.

Thanks a lot man!

Originally Posted by WannaBe

Thanks a lot man!

Please check a modification of Q3.

Originally Posted by aliceinwonderland

Please check a modification of Q3.

modification of Q3?

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.

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 !