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The problem is to prove or disprove that there is a noncyclic abelian group of order $51$.

I don't think such a group exists. Here is a brief outline of my proof:

Assume for a contradiction that there exists a noncyclic abelian group of order $51$.

We know that every element (except the identity) has order $3$ or $17$. Assume that $|a|=3$ and $|b|=17$. Then I managed to prove that the subgroups generated by $a$ and $b$ only intersect at the identity element, from which we can show that $ab$ is a generator of the whole group, so it is cyclic. Contradiction.

So every element (except the identity) has the same order $p$, where $p$ is either $3$ or $17$.

If $p=17$, take $a$ not equal to the identity, and take $b$ not in the subgroup generated by $a$. Then we can prove that $a^kb^l$ where $k,l$ are integers between $0$ and $16$ inclusive are distinct, hence the group has more than $51$ elements, contradiction.

If $p=3$, take $a$ not equal to the identity and take $b$ not in the subgroup generated by $a$. Then we can prove that $a^kb^l$ where $k,l$ are integers betwen $0$ and $2$ inclusive are distinct. This subgroup has $9$ elements so we can find $c$ that's not of the form $a^kb^l$. Then we can prove that $a^kb^lc^m$ where $k,l,m$ are integers betwen $0$ and $2$ inclusive are distinct. Then this subgroup has $27$ elements so we can find $d$ that's not of the form $a^kb^lc^m$. Then we prove that $a^kb^lc^md^n$ where $k,l,m,n$ are integers between $0$ and $2$ inclusive are distinct, this being $81$ elements. Contradiction.

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  • $\begingroup$ Have you learned the Sylow theorems? $\endgroup$
    – coldnumber
    Jul 8, 2015 at 5:31
  • $\begingroup$ I haven't learned the Sylow Theorems. $\endgroup$
    – Cay
    Jul 8, 2015 at 5:31
  • $\begingroup$ If you've learned Lagrange's theorem, then your last paragraph can be shortened. $\;$ $\endgroup$
    – user57159
    Jul 8, 2015 at 6:59
  • $\begingroup$ I have learned Lagrange's Theorem, but I can't see how it can help us shorten the last paragraph. $\endgroup$
    – Cay
    Jul 8, 2015 at 7:53
  • $\begingroup$ Your problem is a particular case of the following problem from Herstein's Book:Let $G$ be a group of order $pq$ ($p,q$ are primes and $p <q$) and $p$ does not divide $q-1$ then $G$ is a cyclic group.Moreover if p divides $q-1$ then there is exactly one nonabelian group of order $pq$.As far as I know proof of this requires a Use of Sylow's Theorem. $\endgroup$ Jul 8, 2015 at 10:05

3 Answers 3

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Using the Sylow theorems would shorten your proof considerably, because from the first Sylow theorem it follows that if a prime divides the order of a group, then the group contains an element of that order. (This eliminates the need to check cases where all elements have order 3 or all have order 17.)

Hence, if $\left|G\right|=51$, then $G$ has an element $a$ of order 3 and an element $b$ of order 17.

Then, as you said, $ab$ has order 51; you can show this directly, without using the proposition for product groups:

Since $\left|G\right|=51$, by Lagrange's theorem, the order of $ab$ divides 51.

Now, since $G$ is abelian, $(ab)^3=a^3b^3=b^3 \neq 1$ because $b$ has order 17, and $(ab)^{17}=a^{17}b^{17}=a^2\neq 1$, because $a$ has order 3.

Hence, $ab$ generates the group, so $G$ is cyclic.

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  • $\begingroup$ Thanks! I like your solution! $\endgroup$
    – Cay
    Jul 8, 2015 at 5:54
  • $\begingroup$ Thanks! I edited because I realized invoking Lagrange is easier than what I was doing to show that the order of $ab$ divides 51. $\endgroup$
    – coldnumber
    Jul 8, 2015 at 6:04
  • $\begingroup$ There is no need for the Sylow theorems here! The result you mention is Cauchy's theorem, much simpler than Sylow's theorems. $\endgroup$ Jul 8, 2015 at 10:27
  • $\begingroup$ @mt_ Interesting! I hadn't heard of it before; Artin has the existence of the elements as a corollary to the first Sylow Theorem. $\endgroup$
    – coldnumber
    Jul 8, 2015 at 10:38
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    $\begingroup$ @ColdNumber, the proof of the Sylow theorems for abelian groups is also very simple! $\endgroup$ Jul 8, 2015 at 22:43
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You're right.

If you know the theorem that classifies finite abelian groups, then the only possible abelian groups of order $51$ are $\mathbb Z/51 \mathbb Z$ which is cyclic and $\mathbb Z/3 \mathbb Z \times \mathbb Z/17 \mathbb Z$ which is also cyclic because $3$ and $17$ are prime and $\gcd(3,17)=1$ so $\mathbb Z/3 \mathbb Z \times \mathbb Z/17 \mathbb Z \simeq \mathbb Z/51 \mathbb Z$.

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Any finite abelian group is isomorphic to a group of the form $$\Bbb Z_{p_1^{a_1}}\times \Bbb Z_{p_2^{a_2}}\times\cdots\times \Bbb Z_{p_n^{a_n}}$$ where $p_i$ are (not necessarily distinct) primes. The order of such a group is $p_1^{a_1}\cdots p_n^{a_n}$. How many ways can this be done for $51$?

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    $\begingroup$ Thanks! So it must be $\mathbb{Z}_{51}$ or $\mathbb{Z}_3 \times \mathbb{Z}_{17}$, both of which are cyclic. I did see that theorem later in my textbook, but I thought I wasn't allowed to use it as this question appeared in an earlier chapter. Does my proof work? $\endgroup$
    – Cay
    Jul 8, 2015 at 5:30
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    $\begingroup$ @Cay Your proof looks very good, except for the parts where you just say "it can be proven that they are all distinct". I believe you, but you would need to actually prove it to get full score in my book, had I been your teacher. $\endgroup$
    – Arthur
    Jul 8, 2015 at 5:40
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    $\begingroup$ @Cay actually, $\mathbb{Z}_{51}$ is isomorphic to $\mathbb{Z}_3\times \mathbb{Z}_{17}$ because gcd(3,17)=1. This is true in general for any integers $m$ and $n$. Maybe you could mess around and try to prove this with a strategy similar to the proof you used in this question. $\endgroup$ Jul 8, 2015 at 5:45
  • $\begingroup$ @mathers101 Yes you're right and I see why this is true now. $\endgroup$
    – Cay
    Jul 8, 2015 at 5:52
  • $\begingroup$ @Arthur Yes if it had been an exam I would have fully written out the reasoning, just wanted to get out the main steps here :) $\endgroup$
    – Cay
    Jul 8, 2015 at 5:52

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