Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
Do you know FTFGAG? – Mr.Guy Mar 5 '13 at 4:21
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
up vote 6 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

share|cite|improve this answer
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.
share|cite|improve this answer
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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