# Thread: Finding all groups of order n up to isomorphism

1. ## Finding all groups of order n up to isomorphism

I'm just getting started with using P-sylow theorems to determine all the groups of order N up to isomorphism.

For example, the question would say: State all the groups of order 12 up to isomorphism. Order of G is 12.

n2 is the number of 2-sylow subgroups of G.
n3 is the number of 3-sylow subgroups of G.

n2 = 1 (mod2) and n2 divides 3 => n2 is 1 or 3
n3 = 1 (mod3) and n3 divides 4 => n3 is 1 or 4

If n3 = 4 then that means there are four 3-sylow subgroups of G. Since all four subgroups have the identity and two unique elements, and all of those 8 elements have order 3 in G. That is one identity, and eight elements of order 3. (n2 must equal 1 in this case because...) A 2-sylow group has order 4 and not counting the identity it adds 3 more elements to the group. Equalling a smashing total of 12 elements. One identity, eight order 3's, and three order 4's.

Q: how do I describe this group? Z2xZ6?
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What about groups of order 28?

Z28 is one
D14 is another

n2=1(mod2) and n2 divides 7 => n2 = 1 or 7
n7=1(mod7) and n7 divides 4 => n7 = 1

Since the 7-sylow group is unique it must be normal in G.

If n2=1 , then the 2-sylow group would be unique and normal in G too. That would make the direct product of Z4xZ7 which is isomorphic to Z28. An abelian group.

If n2=7, then there are seven 2 sylow groups. That makes one identity and 21 elements of order 4. Those 22 elements pluse the 6 elements of order 7 makes a wooping total of 28 elements in G.
Q: How do I describe this group?

I think it's three different semi direct products with three different alphas. That's the function used to multiply the left elements (usually). If the alpha is the identity in Z4 then it would just do nothing and the group would be isomorphic to Z28. If the alpha operates from one of the three other elements that makes three different isomorphic groups??

Is there some website that describes how to get the semi-direct products out of this?

2. Originally Posted by charliex
I'm just getting started with using P-sylow theorems to determine all the groups of order N up to isomorphism.

For example, the question would say: State all the groups of order 12 up to isomorphism. Order of G is 12.

n2 is the number of 2-sylow subgroups of G.
n3 is the number of 3-sylow subgroups of G.

n2 = 1 (mod2) and n2 divides 3 => n2 is 1 or 3
n3 = 1 (mod3) and n3 divides 4 => n3 is 1 or 4

If n3 = 4 then that means there are four 3-sylow subgroups of G. Since all four subgroups have the identity and two unique elements, and all of those 8 elements have order 3 in G. That is one identity, and eight elements of order 3. (n2 must equal 1 in this case because...) A 2-sylow group has order 4 and not counting the identity it adds 3 more elements to the group. Equalling a smashing total of 12 elements. One identity, eight order 3's, and three order 4's.

Q: how do I describe this group? Z2xZ6?
===========================================
To find abelian groups of order 12 up to isomorphism, you need to use the fundamental theorem of finitely generated abelian groups.
Up to isomorphism, there are

$\displaystyle \mathbb{Z}_{12} \cong \mathbb{Z}_4 \times \mathbb{Z}_3$,
$\displaystyle \mathbb{Z}_2 \times \mathbb{Z}_6$.

A group presentation of $\displaystyle \mathbb{Z}_2 \times \mathbb{Z}_6$ is $\displaystyle <r, s | r^2=1, s^6=1, rs=sr>$.

What about groups of order 28?

Z28 is one
D14 is another

n2=1(mod2) and n2 divides 7 => n2 = 1 or 7
n7=1(mod7) and n7 divides 4 => n7 = 1

Since the 7-sylow group is unique it must be normal in G.

If n2=1 , then the 2-sylow group would be unique and normal in G too. That would make the direct product of Z4xZ7 which is isomorphic to Z28. An abelian group.

If n2=7, then there are seven 2 sylow groups. That makes one identity and 21 elements of order 4. Those 22 elements pluse the 6 elements of order 7 makes a wooping total of 28 elements in G.
Q: How do I describe this group?

I think it's three different semi direct products with three different alphas. That's the function used to multiply the left elements (usually). If the alpha is the identity in Z4 then it would just do nothing and the group would be isomorphic to Z28. If the alpha operates from one of the three other elements that makes three different isomorphic groups??

Is there some website that describes how to get the semi-direct products out of this?
We know that there is a unique 7-sylow subgroup in the group of order 28. As you said, if n_2=1, then it simply represents $\displaystyle \mathbb{Z}_{28} \cong (\mathbb{Z}_4 \times \mathbb{Z}_7)$ or $\displaystyle \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_7$.

Now we consider 2-sylow group is not normal. Let H be a normal sylow 7-subgroup such that $\displaystyle H \cong \mathbb{Z}_7$. We see that $\displaystyle \text{Aut (H)} \cong \mathbb{Z}_6$. Now consider a homomorphism, $\displaystyle \phi:K \rightarrow \text{Aut (H)}$ where $\displaystyle K \cong \mathbb{Z}_4$ or $\displaystyle K \cong \mathbb{Z}_2\times \mathbb{Z}_2$. Since $\displaystyle \text{Aut (H)}\cong \mathbb{Z}_6$, there is no subgroup of order 4 in Aut(H). If $\displaystyle \phi$ is not trivial, then $\displaystyle \text{Aut (H)}$ contains a subgroup of order 2. Thus the group of order 28 (up to isomorphism) is either

$\displaystyle C_{14} \times C_2$,

$\displaystyle C_7 \times C_4$ (For a trivial $\displaystyle \phi$),
$\displaystyle C_7 \rtimes_\phi C_4$,
$\displaystyle C_7 \rtimes_\phi (C_2 \times C_2)$, where C_n is a cyclic group of order n, and the subgroup of order 2 in $\displaystyle \text{Aut (H)}$ is $\displaystyle \{1, \sigma\}$ where $\displaystyle \sigma(x)=x^{-1}$ for $\displaystyle x \in H$.

Semidirect product for group classifications.
Group of order pq with pq=21.

3. ## References

Hi--

1) Algebra by M.Artin ....Sylow theorems and applications

2) Abstract algebra by Dummit foote...Classfication of groups of order pq,p^{2}q,...

3) Topics in algebra by I.N.Herstein...This is a truly wonderful book particulary if u are interested in sylow theory..lots of examples and different ways of priving the theorems and challenging excercises....

Also refer the automorphisms section in the same book, as it allows u to construct non abelian groups of order $\displaystyle pq$ where $\displaystyle p,q \ \text{are primes} \ and p \mid (q-1)$.

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# up to isomorphism the number of distinct group of order

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