Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Show that there are two abelian groups of order 108 that have exactly one subgroup of order 3.

$$108 = 2^ 2 \times 3 ^ 3$$

Using the fundamental theorem of finite abelian groups, we have

Possible abelian groups of order 108 are: $\mathbb{Z}_{108}$, $ \mathbb{Z}_4 + \mathbb{Z}_{27}$, $\mathbb{Z}_2+\mathbb{Z}_2+\mathbb{Z}_{27}$, $\mathbb{Z}_4+\mathbb{Z}_9+\mathbb{Z}_3$, $\mathbb{Z}_2+\mathbb{Z}_2+\mathbb{Z}_9+\mathbb{Z}_3$, $\mathbb{Z}_4+\mathbb{Z}_3+\mathbb{Z}_3+\mathbb{Z}_3$, $\mathbb{Z}_2+\mathbb{Z}_2+\mathbb{Z}_3+\mathbb{Z}_3+\mathbb{Z}_3$.

It seems to me that all three $\mathbb{Z}_{108}$, $\mathbb{Z}_4 + \mathbb{Z}_{27}$, $\mathbb{Z}_2+\mathbb{Z}_2+\mathbb{Z}_{27}$, have exactly one subgroup of order 3. Please suggest where I am going wrong.

Is it because $\mathbb{Z}_{108}$ is isomorphic to $\mathbb{Z}_4 + \mathbb{Z}_{27}$?

share|improve this question
2  
Your notation is not quite standard. Are you sure that all of your three groups are different? $Z_4\times Z_{27}$ is different from $Z_2\times Z_2\times Z_{27}$, but maybe one of them is cyclic? –  Someone Sep 28 '11 at 16:02
1  
You should be using $\oplus$ or $\times$, rather than $+$. You can get $\oplus$ by using \oplus. –  Arturo Magidin Sep 28 '11 at 16:07
    
@Someone: Please see my edit it seems to me that z108 and z27+z4 are isomorphic. z4+z27 is cyclic but z2+z2+z27 is not...but both are abelian. –  Tav Sep 28 '11 at 16:08
1  
Why are you counting isomorphic groups separately? –  lhf Sep 28 '11 at 16:09
    
@lhf: I guess you are right cos Z108 and Z4 X Z27 are isomorphic. –  Tav Sep 28 '11 at 16:17
add comment

2 Answers 2

up vote 3 down vote accepted

Yes. For any natural numbers $n$ and $m$ that are coprime (i.e. have no common factors), $$\mathbb{Z}_{mn}\cong\mathbb{Z}_m\times \mathbb{Z}_n$$ so your list of all abelian groups of order 108 (up to isomorphism) actually has a duplicate: $\mathbb{Z}_{108}$ and $\mathbb{Z}_4\times\mathbb{Z}_{27}$ are isomorphic, and so should not be counted as different for these purposes.

So, up to isomorphism, there are only two abelian groups of order 108 with exactly one subgroup of order 3, namely $\mathbb{Z}_{108}$ and $\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_{27}$.

share|improve this answer
    
Thanks for your help. –  Tav Sep 28 '11 at 16:42
add comment

Note that $\mathbb{Z}_4\oplus\mathbb{Z}_{27}\cong \mathbb{Z}_{108}$. In general, if $p$ and $q$ are relatively prime, then $\mathbb{Z}_p\oplus\mathbb{Z}_q \cong \mathbb{Z}_{pq}$.

So you should not list them separately: you are listing isomorphism types, not different ways of writing them. So, yes, the $3$-part of $A$ must be cyclic, as otherwise it has at least two subgroups of order $3$, which means $A$ must be isomorphic to either $\mathbb{Z}_{4}\oplus\mathbb{Z}_{27}$, or to $\mathbb{Z}_2\oplus\mathbb{Z}_2\oplus\mathbb{Z}_{27}$.

(You don't need to check all possibilities: just remember that $A$ is the direct sum of its $p$-parts, so you only need to worry about the $3$-part; this is either $\mathbb{Z}_{27}$, $\mathbb{Z}_3\oplus\mathbb{Z}_{9}$, or $\mathbb{Z}_3\oplus\mathbb{Z}_3\oplus\mathbb{Z}_3$).

share|improve this answer
    
Please explain what do you mean by " 3-part" . –  Tav Sep 28 '11 at 16:40
1  
@Tav: Same idea: since an abelian group of order $108$ is the sum of its $2$-part and its $3$-part, and a subgroup of order $3$ is necessarily a subgroup of the $3$-part, you are only placing constraints on the $3$-part. The $3$-part has order $27$, so the possible $3$-parts are $\mathbb{Z}_{27}$, which has a unique subgroup of order $3$; $\mathbb{Z}_3\oplus\mathbb{Z}_9$, which has four subgroups of order $3$; and $\mathbb{Z}_3\oplus\mathbb{Z}_3\oplus\mathbb{Z}_3$, which has thirteen subgroup of order $3$. (cont) –  Arturo Magidin Sep 28 '11 at 18:21
1  
@Tav: (cont) So the $3$-part must be $\mathbb{Z}_3\oplus\mathbb{Z}_9$. The $2$-part can be anything (it doesn't affect the number of subgroups of order $3$) and must be of order $4$, so it is either $\mathbb{Z}_4$ or $\mathbb{Z}_2\oplus\mathbb{Z}_2$. So the two possibilities are $\mathbb{Z}_4\oplus(\mathbb{Z}_3\oplus\mathbb{Z}_9)$ and $(\mathbb{Z}_2\oplus\mathbb{Z}_2)\oplus(\mathbb{Z}_3\oplus\mathbb{Z}_9)$. –  Arturo Magidin Sep 28 '11 at 18:22
1  
@Tav: To see that $\mathbb{Z}_{27}$ has only one subgroup of order $3$, note that it is cyclic, so it has a unique subgroup of order $d$ for every $d$ that divides $27$. To see that $\mathbb{Z}_3\oplus\mathbb{Z}_3\oplus\mathbb{Z}_3$ has 13 subgroups, note that every element other than the identity generates a cyclic group of order $3$, but that each element and its inverse generate the same subgroup. So you get $26/2=13$ distinct subgroups. Of course, groups of order $3$ must be cyclic. (cont) –  Arturo Magidin Sep 28 '11 at 18:25
1  
@Tav: (cont) To see that $\mathbb{Z}_3\oplus\mathbb{Z}_9$ has three subgroups, note that $(a,b)$ is of order $3$ if and only if $a$ is trivial or of order $3$ (three possibilities), $b$ is trivial of order $3$ (three possibilities), and $a$ and $b$ are not both trivial (one case). That gives $9-1=8$ possible subgroups; but again, an element and its inverse generate the same cyclic group of order $3$, so you are counting each subgroup twice, giving you $4$ distinct subgroups. –  Arturo Magidin Sep 28 '11 at 18:26
show 5 more comments

Your Answer

 
discard

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.