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!


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)

  • $\begingroup$ 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

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