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Show that all abelian groups of order 21 and 35 are cyclic. I have no idea on how to start. Can anyone give some hints?

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Do you know FTFGAG? –  Mr.Guy Mar 5 '13 at 4:21
2  
Do you mean fundamental theorem of finite generated abelian group ? I din learn this theorem. –  Idonknow Mar 5 '13 at 4:22
    
It's worth noting that the abelian requirement is superfluous when the order is 35. See here. –  JSchlather Mar 5 '13 at 5:08

3 Answers 3

up vote 4 down vote accepted

All you need is


With this, you are equipped to conclude what you need for every abelian group of order 35, 21, respectively.

Note that $\mathbb Z_5 \times \mathbb Z_7 \cong \mathbb Z_{35}$, because $\gcd(5, 7) = 1$. And it follows that $\mathbb Z_{35}$ is cyclic.

Similarly, you can work with $\mathbb Z_3 \times \mathbb Z_7 \cong \mathbb Z_{21}$ because the $\gcd(3, 7) = 1$. And hence, $\mathbb Z_{21}$ is cyclic

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Does this look familiar, Idonknow? –  amWhy Mar 5 '13 at 4:41
    
I don know the theorem but I do know the second statement. We prove the second statement by defining a map and prove the map is isomorphism right ? –  Idonknow Mar 5 '13 at 4:43
    
Exactly. That's really all you need for this, the second statement. Check out the link, but don't try to learn it all in a day. –  amWhy Mar 5 '13 at 4:44
    
Feel free to follow up in a comment here if you want to check anything more out, as you write up the proof/solution. –  amWhy Mar 5 '13 at 4:50
    
:^) nice Amy... –  Babak S. Mar 5 '13 at 12:58

You can do this with elementary tools. Here is a possible plan for an abelian group $G$ of order 21.

  1. By Lagrange's theorem, the order of any element is one of 1, 3, 7 or 21.
  2. Take an arbitrary element $a \neq 1$. Its order is either 3, 7 or 21. Suppose it is 3 (the case when it is 7 is similar, and if it is 21 then we are done).
  3. The quotient group $G/\langle a \rangle $ has order $7$, so in fact $G/\langle a \rangle \cong \mathbb{Z}_7$. Every element in $\mathbb{Z}_7$ except the identity has order $7$. Then every element in $G - \langle a \rangle$ has order that is divisible by $7$. Then there is an element $b \in G$ of order $7$.
  4. So, we have elements $a$ and $b$ of orders $3$ and $7$. It is easy to see that then $G$ is a direct product $G = \langle a \rangle \times \langle b \rangle$, so $G \cong \mathbb{Z}_3 \times \mathbb{Z}_7$. It is in fact cyclic, qed.
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If you remove "suppose $G$ is not cyclic" at the top and ",so this is a contradiction." you realize this is really a direct proof. Lagrange's theorem doesn't rely on the group being cyclic, so your proof works out. –  Pedro Tamaroff Mar 5 '13 at 5:19
    
@Peter Yes, it does work out. It didn't work out this way in my original approach, so the proof by contradiction is here for historic reasons ) –  Dan Shved Mar 5 '13 at 5:21
    
My sole point is that it is kinda meaningless to include those phrases, since this is not a proof by contradiction. Moreover, it might give rise to confusion. –  Pedro Tamaroff Mar 5 '13 at 5:23
    
@Peter I see your point, but if I try to rewrite this as a direct proof, then steps 2 and 3 will become messy. In step 2 i'll have to add something like "and if the order of $a$ is 21, we are done". And I'll have to add a similar remark about the order of $b$ in step 3. This will probably look messy. Could you maybe suggest a way to avoid that? –  Dan Shved Mar 5 '13 at 5:26
    
Oh, now I see the catch, my bad. But still I will try and think something. –  Pedro Tamaroff Mar 5 '13 at 5:30

HINT: There are only 2 possible abelian groups of order 21: $\mathbb{Z}_{21}$ and $\mathbb{Z}_3\times\mathbb{Z}_7$. You can show that the latter is cyclic by exhibiting a generator (it's probably the first thing you'll think of); in fact these groups are isomorphic.

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