2

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?

amWhy
  • 209,954
Idonknow
  • 15,643

3 Answers3

7

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

amWhy
  • 209,954
6

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.
Dan Shved
  • 15,862
  • 39
  • 55
  • 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 Mar 05 '13 at 05:19
  • @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 05 '13 at 05:21
  • 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 Mar 05 '13 at 05:23
  • @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 05 '13 at 05:26
  • Oh, now I see the catch, my bad. But still I will try and think something. – Pedro Mar 05 '13 at 05:30
  • @Peter I've rewritten it as a direct proof, and it looks better, even with those additional remarks. So, thanks for suggesting an improvement to the answer :-) – Dan Shved Mar 05 '13 at 05:34
  • And if it's not clear to anyone this exact proof goes through if you replace $3$ with $5$ and $21$ with $35$. Or more notably if you replace $3$ and $7$ with any two distinct primes. – JSchlather Mar 05 '13 at 06:03
  • Does this proof actually rely on it being abelian? If it does could you please clarify which step (or part thereof) doesn't hold if it's non-abelian – Ainsley Pullen Jul 22 '15 at 05:46
  • @Alex it relies on abelianness in step 4, in the "easy to see" part. One has to consider products like $a^m b^n$ and use the fact that $(a^m b^n)(a^{m'}b^{n'}) = a^{m + m'} b^{n + n'}$, which is true due to the abelianness. Without abelianness one can construct a semidirect product of $\mathbb{Z}_3$ and $\mathbb{Z}_7$ that is not isomorphic to $\mathbb{Z}_3 \times \mathbb{Z}_7$. – Dan Shved Jul 22 '15 at 22:52
2

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.

Chris Brooks
  • 7,424