hi! I really would appreciate if someone can solve these questions..
thanks a lot!

1. let G be a group from order 297. prove that there is a subgroup K of G wich
the center of G/K is not Trivial.

2. look at A5 :
a. how many numbers from order 3 are in A5?
b. how many 3-Sylow groups are in A5?
c. let H be 2-Sylow subgroup of A5. prove that H abelian and find wich abelian group
H is isomorphic.

3. prove or disprove:
a.there is no group G and a normal subgroup H of G (when H is not G or {1G}) so that G/H isomorphic to G.
b. there is no group G and subgroup H [so that H is not G] so that G
isomorphic to H.
c. for all n natural there is a group G and a subgroup H
[wich both are Independent in n] so that rank(G)=2 and rank(H)=n
d. for all G group from order n, there are no m<n (when m belongs to N)
so that there is monomorphism G->Sm.
e. for all group G , H subgroup of G so that: Z(H) is a subgroup of Z(G).

2. Originally Posted by themanandthe
hi! I really would appreciate if someone can solve these questions..
thanks a lot!

1. let G be a group from order 297. prove that there is a subgroup K of G wich
the center of G/K is not Trivial.
Using a simple application of Sylow's theorems, you want to show that there is only one Sylow 11-subgroup. Why is this sufficient?

Originally Posted by themanandthe
2. look at A5 :
a. how many numbers from order 3 are in A5?
b. how many 3-Sylow groups are in A5?
c. let H be 2-Sylow subgroup of A5. prove that H abelian and find wich abelian group
H is isomorphic.
For (a) I ask must ask you a question...what does an element of order 3 in $\displaystyle A_5$ look like (in cycle notation)? Can you count how many elements have this form?

For (b), you know that there are $\displaystyle n$' elements of order 3. Each of these must appear in some Sylow 3-subgroup of order 3, and so they must appear in pairs in this subgroup. So your answer is $\displaystyle n/2$ (do you understand why this is)?

You should check that this answer is valid by plugging in Sylow's Theorems (they do not give you the actual answer in this case, just a couple of possible answers).

For (c), what is the order of a Sylow 2-subgroup? I shall call this number $\displaystyle m$'. Now, can you think of a well-known result about groups of order m'? For the next bit, every group of order $\displaystyle m$ is a Sylow 2-subgroup (why?), and as all such groups are conjugate they are all isomorphic. Therefore, to finish the question you need to find a subgroup (any subgroup!) of $\displaystyle A_5$ of order $\displaystyle m$'. What does this group look like?

Originally Posted by themanandthe
3. prove or disprove:
a.there is no group G and a normal subgroup H of G (when H is not G or {1G}) so that G/H isomorphic to G.
b. there is no group G and subgroup H [so that H is not G] so that G
isomorphic to H.
c. for all n natural there is a group G and a subgroup H
[wich both are Independent in n] so that rank(G)=2 and rank(H)=n
d. for all G group from order n, there are no m<n (when m belongs to N)
so that there is monomorphism G->Sm.
e. for all group G , H subgroup of G so that: Z(H) is a subgroup of Z(G).
Hmm...(a) and (b) are actually false, but I suspect you are only looking at finite groups and so are disproved by simple order arguments. (For (b), a counter-example is the integers...I can't think of a counter-example for (a) off-hand EDIT: Take the infinte cross-product of $\displaystyle \mathbb{Z}$ with itself. This is an infinitely generated abelian group. Then, quotient out one of the $\displaystyle \mathbb{Z}$'s...).

As a hint for (c), Cayley's Theorem'.

For (d), I believe your first counter-example comes at order 6. However, you should note that $\displaystyle S_i \leq S_j$ for $\displaystyle i \leq j$. So...

And I will leave you to solve (e). You proof should start Let $\displaystyle g \not\in Z(G)$...' (again, it is false).

3. Swlabr - wow! thanks a lot for your help!!! really!

i'm kindda new at this so i'm trying to understand what you have wrote but still, i would really appreciate if you could simplfy a bit more for me..

1. Using a simple application of Sylow's theorems - how do i use it?
2. (a) - i really dont know.. sorry about that.
2.(b) - do you understand why this is)? now i dont..
2.(c) Therefore, to finish the question you need to find a subgroup (any subgroup!) of of order '. What does this group look like? can u explain? i can't get it right..

3.(a) Then, quotient out one of the 's - how exactly i do that?
3.(c) - Cayley's Theorem' -???? i can't see it..
3(e) - sorry but again i dont understand how..

thanks a lot! sorry for all the question. just starting to learn this subject..

4. Originally Posted by themanandthe
Swlabr - wow! thanks a lot for your help!!! really!

i'm kindda new at this so i'm trying to understand what you have wrote but still, i would really appreciate if you could simplfy a bit more for me..

1. Using a simple application of Sylow's theorems - how do i use it?
Have you covered Sylow's Theorems? If so, go back to where it was covered. There should be a similar example.

Essentially, Sylow's Theorems are three theorems.

-The first says if $\displaystyle |G| = p^nm$ and $\displaystyle gcd(p, m)=1$ with $\displaystyle p$ prime then there exists a subgroup of order $\displaystyle p^n$ in $\displaystyle G$'. Such a subgroup is called a Sylow $\displaystyle p$-subgroup.

-The second says that all Sylow $\displaystyle p$-subgroups are conjugate for a fixed $\displaystyle p$. Thus, they are always isomorphic, and if there is only one it is normal.

-The third gives some information on the number of Sylow $\displaystyle p$-subgroups for a fixed $\displaystyle p$, denoted $\displaystyle n_p$. Specifically,

$\displaystyle n_p | m$
$\displaystyle n_p \equiv 1 \text{ mod }p$
and a third condition to do with the normaliser.

You want to use the first two of those conditions to find the number of Sylow 11-subgroups. What does this mean? How can you apply this to your question (you will need to use the fact that the center of a p-group is non-trivial for p prime.)

Originally Posted by themanandthe
2. (a) - i really dont know.. sorry about that.
2.(b) - do you understand why this is)? now i dont..
2.(c) Therefore, to finish the question you need to find a subgroup (any subgroup!) of of order `'. What does this group look like? can u explain? i can't get it right..
Okay, first things first. What does an element of order 3 in $\displaystyle S_5$ look like? Which of these are in $\displaystyle A_5$?

Originally Posted by themanandthe

3.(a) Then, quotient out one of the 's - how exactly i do that?
3.(c) - Cayley's Theorem' -???? i can't see it..
3(e) - sorry but again i dont understand how..

thanks a lot! sorry for all the question. just starting to learn this subject..
Ignore what I said for 3(a) - just use counting arguments to look at the finite case.

For 3(c), do you know what Cayley's Theorem says? How many generators does $\displaystyle S_n$ have?

For 3(e), you have an element outwith the center. So, you want to put this in the center of a subgroup....I will, however, leave you to ponder/solve it.