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I have an abelian group $G$ and two cyclic subgroups of orders $p,q$, $( p,q)=1$ and I need to show I have a subgroup of order $pq$.

Non-abelian $p$-group; abelian subgroups of index $p$Do finite $p$-groups have subgroups of all possible orders containing a given non-normal subgroup?Group of order $135$ abelian and not cyclicLet $G$ be an abelian group;$H$, $K$ are finite cyclic subgroups. Show that $G$ contains a cyclic subgroup of order $lcm(r, s)$.Show by Lagrange's theorem that a group $G$ of order $27$ should have a subgroup of order $3$.All groups of order 10 have a proper normal subgroupprove that non cyclic group has two cyclic subgroups of different ordersFinding cyclic subgroup generators given orders$p$-group as product of two normal cyclic subgroupsIf every proper subgroup of a finite abelian group $G$ is cyclic, then $G$ is cyclic.

1 Answer
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3

$begingroup$

Hint: If $x$ has order $p$ and $y$ has order $q$, what is the order of $xy$?

In fact, we do not even need that the given subgroups are cyclic. In most textbooks (e.g. Dummit and Foote), you will find the more general proposition that if $HK = KH$ for subgroups $H,K<G$, then
$$
|HK|=frac{|H||K|}{|Hcap K|}.
$$
The conditions on $H$ and $K$ are trivially satisfied since $G$ is abelian. Also, $Hcap K$ is a subgroup of both $H$ and $K$, so its order must divide $|H|$ and $|K|$. But, when $|H|$ and $|K|$ are relatively prime, this implies $|Hcap K |=1$.

(As pointed out by @Tobias in his comment, the identity holds without the condition that $HK = KH$; the condition ensures that $HK$ is a subgroup of $G$.)

share|cite|improve this answer

edited Mar 15 at 12:15

answered Mar 15 at 8:21

o.h.o.h.

6917

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Actually, the condition that $HK = KH$ is only needed for the product to be a subgroup. The size of the product will always be given by that formula.
  $endgroup$
  – Tobias Kildetoft
  Mar 15 at 10:26
  

  
  
  
  

add a comment | 

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3

$begingroup$

Hint: If $x$ has order $p$ and $y$ has order $q$, what is the order of $xy$?

In fact, we do not even need that the given subgroups are cyclic. In most textbooks (e.g. Dummit and Foote), you will find the more general proposition that if $HK = KH$ for subgroups $H,K<G$, then
$$
|HK|=frac{|H||K|}{|Hcap K|}.
$$
The conditions on $H$ and $K$ are trivially satisfied since $G$ is abelian. Also, $Hcap K$ is a subgroup of both $H$ and $K$, so its order must divide $|H|$ and $|K|$. But, when $|H|$ and $|K|$ are relatively prime, this implies $|Hcap K |=1$.

(As pointed out by @Tobias in his comment, the identity holds without the condition that $HK = KH$; the condition ensures that $HK$ is a subgroup of $G$.)

share|cite|improve this answer

edited Mar 15 at 12:15

answered Mar 15 at 8:21

o.h.o.h.

6917

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Actually, the condition that $HK = KH$ is only needed for the product to be a subgroup. The size of the product will always be given by that formula.
  $endgroup$
  – Tobias Kildetoft
  Mar 15 at 10:26
  

  
  
  
  

add a comment | 

3

$begingroup$

Hint: If $x$ has order $p$ and $y$ has order $q$, what is the order of $xy$?

In fact, we do not even need that the given subgroups are cyclic. In most textbooks (e.g. Dummit and Foote), you will find the more general proposition that if $HK = KH$ for subgroups $H,K<G$, then
$$
|HK|=frac{|H||K|}{|Hcap K|}.
$$
The conditions on $H$ and $K$ are trivially satisfied since $G$ is abelian. Also, $Hcap K$ is a subgroup of both $H$ and $K$, so its order must divide $|H|$ and $|K|$. But, when $|H|$ and $|K|$ are relatively prime, this implies $|Hcap K |=1$.

(As pointed out by @Tobias in his comment, the identity holds without the condition that $HK = KH$; the condition ensures that $HK$ is a subgroup of $G$.)

share|cite|improve this answer

edited Mar 15 at 12:15

answered Mar 15 at 8:21

o.h.o.h.

6917

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Actually, the condition that $HK = KH$ is only needed for the product to be a subgroup. The size of the product will always be given by that formula.
  $endgroup$
  – Tobias Kildetoft
  Mar 15 at 10:26
  

  
  
  
  

add a comment | 

$begingroup$

Hint: If $x$ has order $p$ and $y$ has order $q$, what is the order of $xy$?

In fact, we do not even need that the given subgroups are cyclic. In most textbooks (e.g. Dummit and Foote), you will find the more general proposition that if $HK = KH$ for subgroups $H,K<G$, then
$$
|HK|=frac{|H||K|}{|Hcap K|}.
$$
The conditions on $H$ and $K$ are trivially satisfied since $G$ is abelian. Also, $Hcap K$ is a subgroup of both $H$ and $K$, so its order must divide $|H|$ and $|K|$. But, when $|H|$ and $|K|$ are relatively prime, this implies $|Hcap K |=1$.

(As pointed out by @Tobias in his comment, the identity holds without the condition that $HK = KH$; the condition ensures that $HK$ is a subgroup of $G$.)

share|cite|improve this answer

edited Mar 15 at 12:15

answered Mar 15 at 8:21

o.h.o.h.

6917

$endgroup$

share|cite|improve this answer

edited Mar 15 at 12:15

answered Mar 15 at 8:21

o.h.o.h.

6917

answered Mar 15 at 8:21

o.h.o.h.

6917

answered Mar 15 at 8:21

o.h.o.h.

6917