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I've tried to check if $U_{14}$ and $U_{18}$ are isomorphic to each other , so first :

$U_{18}=\{1,5,7,11,13,17\}$

$U_{14}=\{1,3,5,9,11,13\}$

Both of them of order 6, cyclic, and probably abelian (I'm not quite sure yet regarding the abelian though, I need to do a little more digging).

Questions :
1. Why can I say that $U_{18}$ and $U_{14}$ are isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_3$?
2. How can I know if $U_{18}$ and $U_{14}$ are abelian ?

Thanks

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Do you mean to say, you know the groups are cyclic, but not sure if they are abelian? –  user21436 Mar 9 '12 at 3:59
4  
Cyclic groups are always abelian; a group element (in particular, a generator for a cyclic group) commutes with itself. –  Dylan Moreland Mar 9 '12 at 4:02
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How do you define $U_{14}$ and $U_{18}$? It looks like these are the multiplicative groups ${(\mathbb{Z}/14)}^*$ and ${(\mathbb{Z}/18})^*$ (multiplicative units mod 14 and 18). –  Brett Frankel Mar 9 '12 at 4:06
    
Hint: Any finite cyclic group of order $n$ is isomorphic to $(\mathbb{Z}_n, +)$. Furthermore, $\mathbb{Z}_{mn}$ is isomorphic to $\mathbb{Z}_m \times \mathbb{Z}_n$ when $\gcd(m,n) = 1$. –  JavaMan Mar 9 '12 at 4:35

3 Answers 3

up vote 1 down vote accepted

Well, I am afraid if I will kill your learning process by giving away complete answers. But, nonetheless. I believe I should emphasize that you'll have to give all these questions a thought before you think you can post them here. For one, I am sure you can answer all your questions yourself.

Here are the hints.

  1. A group $G$ is said to be direct product if and only if there exists subgroups $H$ and $K$ such that,

    a) $H$ and $K$ are normal in $G$.

    b) $H \cap K= \{e_G\}$

    c) $HK=G$

If (a), (b) and (c) are met, we write, $G=H \times K$

Now, can you find an element of order $2$ in your groups? of order $3$? Can the groups they generate intersect non-trivially $^\dagger$ ? What is the order of the product of the groups generated by these elements? So, what can you conclude?

  • Direct product of abelian groups is abelian. Cyclic groups are also abelian.

  • (Additional Exercise) There are only two groups of order $6$ upto isomorphism. What are they? Can you prove that?


$\dagger$ Non-trivially means, in a "set" that contains some elements other than identity. I am not sure if you have proved, if $H$ and $K$ are subgroups of $G$ such that $(|H|,|K|)=1$, then $H \cap K= \{e_G\}$. Further, the intersection of subgroups is a subgroup and this means I can replace "set" by subgroup. (The lasst fact is required to prove the previous fact about trivial intersection.)

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In 1, The subgroups $H$ and $K$ must both be normal in $G$. –  Dane Mar 9 '12 at 4:14
    
@Kannappan Sampath: Okay ,for U14: element "13" has order 2. For U18 element "17" has order 2. Now what does it mean exactly ? first that both U18 & U14 are cyclic for sure ,since they have only one element of order 2. But what else ? thanks !! –  ron Mar 9 '12 at 6:12
    
The point I am trying to push through is you can prove the direct product thing without having to classify all groups of order $6$. So, proceed with the next step. Produce an element of order $3$ and prove that the rroup generated by the elements you had produced don't intersect non-trivially. So,.... –  user21436 Mar 9 '12 at 8:19
    
@ron Further, having a unique element of order $2$ does not guarantee cyclicness of the group. A finite group is cyclic iff for each divisor of the order of the group, there is an unique subgroup of that order. For a counter example look at $\Bbb Z_3 \times \Bbb Z_3 \times \Bbb Z_2$. –  user21436 Mar 9 '12 at 8:21

There are only two groups of order 6: $C_6$ (cyclic) and $S_6$ (non-abelian). Since your groups are abelian, they must be isomorphic to $C_6$, which is isomorphic to $Z_2 \times Z_3$.

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I don't understand , If Euler group U14 and U18 has 6 elements , it doesn't mean that their order is 6 ? –  ron Mar 9 '12 at 4:22
    
Okay I got it ... :) –  ron Mar 9 '12 at 4:26

Hint $\ $ Both are $\rm\:\{(-1)^k (-5)^n\},\ \ k\in \mathbb Z/2, \ \ n\in \mathbb Z/3$

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