0
$\begingroup$

Suppose that G is an abelian group of order 66. Show that G is a cyclic group and show that there are proper subgroups $H$ and $K$ of $H$ such that $G\cong G/H \times G/K$

Now, for this question, I would like to use the results of a previous question that I asked; Let $\theta : G \rightarrow G/H \times G/K$ defined by $g\rightarrow (gH)(gK)$, show that $G/(H\cap K) \cong$ to a subgroup of $G/H \times G/K$.

For the first part, we have that $66=2^13^1 11^1$. Therefore, by using Cauchy's theorem, G has elements $x,y,z$ of order $2,3$ and $11$ respectively. These commute and are relatively prime, so the order of $xyz$ is $66$ and G is cyclic. Please let me know if this part follows, I have only seen this being done with two numbers!

Now, let us have the subgroups $K=\langle x, y \rangle$ and $H=\langle z \rangle$. I don't really know where to go from here!

$\endgroup$
0
$\begingroup$

For your first question: relatively prime is not enough. In fact, the following is true:

if $g_1,\ldots,g_r\in G$ pairwise commute, and have pairwise coprime orders, then the order of $g_1\cdots g_r$ is the product of the orders of $g_1,\ldots,g_r$.

I let you prove this by induction on $r$, starting from the case $r=2$ you seem to know. You might want to find a counterample if the orders are globally relatively prime but not pairwise comprime.

For your last question, I suggest you compute $\vert K\vert$ and $\vert H\vert$ and see that they are coprime. What can you say about the intersection of two subgroups with coprime orders ? (Hint: Lagrange 's theorem)

$\endgroup$
  • $\begingroup$ The intersection of two groups with coprime orders is trivial. Therefore, |K|=6 and |H|=11, but from here I don't know where to go! $\endgroup$ – Catherine Drysdale Jun 1 '17 at 10:01
  • $\begingroup$ Well, your map $\theta$ is then injective (its kernel is $H\cap K$). Compare now the orders of the groups $G$ and $G/H\times G/K$. $\endgroup$ – GreginGre Jun 1 '17 at 10:04
  • $\begingroup$ The orders will be the same, but this must be the case for the injectivity. I wasw wondering how do we actually define the isomorphism. x $\endgroup$ – Catherine Drysdale Jun 1 '17 at 10:17
  • $\begingroup$ "The orders will be the same, but this must be the case for the injectivity". Certainly not. If a map $E\to F$ between to finite set is injective, then $\vert E\vert \leq \vert F\vert$. You don't need these two numbers to be equal. Well, $\theta$ is an injective group morphism between two groups having same cardinality. So ? (what can you say of an injective map between two finite sets with the same number of elements?) $\endgroup$ – GreginGre Jun 1 '17 at 10:40
  • $\begingroup$ Ah, it is bijective therefore we have a homomorphism. :) $\endgroup$ – Catherine Drysdale Jun 1 '17 at 10:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.