Classifying Groups

• Dec 2nd 2011, 03:03 PM
dwsmith
Classifying Groups
In my course, we didn't have time to cover classifying groups but we are expect to know it. I need some help in how this is done.

$|G|=70=2\times 5\times 7$

Let $n_i$ be the number of $\text{Syl}_p(G)$ for $i=2,5,7$

$n_i\equiv 1 \ (\text{mod} \ i) \ \text{and} \ n_i|\frac{70}{i}$

$n_2=1,5,7,35$
$n_5=1$
$n_7=1$

So G is not simple and the Sylow 5 and 7 are normal in G. Now how do I classify the groups of order 70?
• Dec 2nd 2011, 06:58 PM
Drexel28
Re: Classifying Groups
Quote:

Originally Posted by dwsmith
In my course, we didn't have time to cover classifying groups but we are expect to know it. I need some help in how this is done.

$|G|=70=2\times 5\times 7$

Let $n_i$ be the number of $\text{Syl}_p(G)$ for $i=2,5,7$

$n_i\equiv 1 \ (\text{mod} \ i) \ \text{and} \ n_i|\frac{70}{i}$

$n_2=1,5,7,35$
$n_5=1$
$n_7=1$

So G is not simple and the Sylow 5 and 7 are normal in G. Now how do I classify the groups of order 70?

This is not a nice problem. Let $P,Q$ be the Sylow $5,7$-subgroups of $G$ respectively. Note that they are both normal, and so their product $N$ is normal in $G$ and of of order $35$. Now, note that trivially if $T$ denotes the Sylow $2$ subgroup of $G$ then $N\cap T=\{1\}$, $NT=G$ where both follow from order arguments. Thus, $G\cong T\rtimes_\varphi N$. Thus, we are left to classify homomorphisms $\varphi:T\to\text{Aut}(N)$ which induce different semidirect products. But, note that $\text{Aut}(N)\cong \text{Aut}(\mathbb{Z}_5)\times\text{Aut}(\mathbb{Z }_7)\cong \mathbb{Z}_4\times\mathbb{Z}_6$, since any group of order $35$ is cyclic, and so isomorphic to $\mathbb{Z}_5\times\mathbb{Z}_7$, $\text{Aut}$ splits accross the product of cyclic groups, and $\text{Aut}(\mathbb{Z}_p)\cong\mathbb{Z}_{p-1}$ for every prime $p$ . And, since $T\cong\mathbb{Z}_2$ we can count the number of possible homomorphisms $T\to\text{Aut}(N)$ by counting the number of possible homomorphisms $\mathbb{Z}_2\to\mathbb{Z}_4\times\mathbb{Z}_6$. But, this clearly corresponds to the number of elements of order two in $\mathbb{Z}_4\times\mathbb{Z}_6$ which is four. Thus, there are at most four homomorphisms $\varphi:T\to\text{Aut}(N)$ and so at most four non-isomorphic groups of order $70$. That said, the groups $\mathbb{Z}_{70}$ (the only possible cyclic group), $D_{7}\times \mathbb{Z}_5$, $D_5\times\mathbb{Z}_7$, and $D_{35}$ are four non-isomorphic groups of $35$. Thus, these are all the possible groups of order $70$.
• Dec 2nd 2011, 07:14 PM
dwsmith
Re: Classifying Groups
Why are you forming the semi-direct product of T and Aut(N)?
• Dec 2nd 2011, 07:18 PM
Drexel28
Re: Classifying Groups
Quote:

Originally Posted by dwsmith
Why are you forming the semi-direct product of T and Aut(N)?

Look here and see that all the necessary conditions for $G$ to be a semidirect product of $T$ by $N$ are satisfied. This is a common technique for classifying groups of a given order--reduce the problem to classifying semidirect products of certain subgroups.
• Dec 3rd 2011, 01:03 AM
Deveno
Re: Classifying Groups
i believe he meant:

$G\cong N\rtimes_\varphi T$, for which we need $\varphi \in \text{Hom}_{\text{\bf{Grp}}}(T,\text{Aut}(N))$

the idea is, if you have two subgroups of G, H and N, but only one of them is normal, you can still use G = HN to "decompose" G (as long as H∩N = {e}).

the classic example is $D_{2n}$ the dihedral group with 2n elements. here, $\langle r \rangle \cong \mathbb{Z}_n$ and $\langle s \rangle \cong \mathbb{Z}_2$.

now we have two obvious choices for a homomorphism:

$\varphi:\mathbb{Z}_2 \to \text{Aut}(\mathbb{Z}_n)$, one is:

$\varphi(k) = \text{id}_{\mathbb{Z}_n}$ for k = 0,1 (the trivial homomorphism, where φ maps everything to the identity automorphism)

this leads to the direct product $\mathbb{Z}_n \times \mathbb{Z}_2$ (which is abelian).

the other "obvious" choice is:

$\varphi(0) = \text{id}_{\mathbb{Z}_n}\ (k \to k)$
$\varphi(1) = (k \to -k)$ (the inversion automorphism).

this leads to $\mathbb{Z}_n \rtimes \mathbb{Z}_2$, which has the multiplication:

$(k,m)*(k',m') = (k+(-1)^mk',m+m')$

in particular:

$(0,1)*(1,0) = (-1,1) = (-1,0)*(0,1)$ which is the "semi-direct product" form of the relation: $sr = r^{-1}s$.

it turns out that this is equivalent to letting H act on N by conjugation (which works because/when N is normal). this tells us quite a bit about G from knowing things about H and N.

ther is one caveat, however. it may be that G = HN (where H and N have trivial intersection), with N normal in G, and that G' = H'N' (with N' normal in G', and H' and N' have trivial intersection), and further: that H ≅ H', N ≅ N', but G is NOT isomorphic to G'. for example:

$\mathbb{Z}_n \times \mathbb{Z}_2 \not \cong D_{2n}$

in general, groups of order $2^n$ are quite difficult to classify. for example there are 49,487,365,422 non-isomorphic groups of order 1024 (i think there are 51 non-isomorphic groups of order 32, classifying groups is a "hard" problem (not impossible, though)).