Partitioning into Isomorphism Classes: Exam Q&A

[Zorba]

Just had an exam there, one of the questions was

Partition the list of groups below into isomorphism classes

1.$$\mathbb{Z}_8$$

2.$$\mathbb{Z}_8^*$$ (elements of Z_8 relatively prime to 8)

3.$$\mathbb{Z}_4 \times \mathbb{Z}_2$$

4.$$\mathbb{Z}_{14} \times \mathbb{Z}_5$$

5.$$\mathbb{Z}_{10} \times \mathbb{Z}_7$$

6.$$D_{70}$$ Symmetries of a regular 35-gon.

7.$$\mathcal{P}\{1,2\}$$ with symmetric difference as operation.

8.Fourth roots of unity

9.$$\langle (12)(34),(13)(24)\rangle \le S_4$$

10. The set generated by the matrix representations of the quaternions.


I don't feel confident about my answers, but I said that 2+9, 4+5, and the rest in separate classes.
I grouped them first by order and then mainly examined element orders, but I'm pretty sure I left some out. Any thoughts?

 

[micromass]

Hi Zorba!

I hope your exam was a good one. Here are my thoughts on the matter:

Just had an exam there, one of the questions was

Partition the list of groups below into isomorphism classes

1.$$\mathbb{Z}_8$$

Cyclic group with 8 elements.

2.$$\mathbb{Z}_8^*$$ (elements of Z_8 relatively prime to 8)

This group contains the elements {1,3,5,7}, thus the group has order 4. The only groups of order 4 are $\mathbb{Z}_4$ and $\mathbb{Z}_2\times\mathbb{Z}_2$. Checking the orders of all elements gives us that all elements have order at most 2. Thus the group is $\mathbb{Z}_2\times \mathbb{Z}_2$.

3.$$\mathbb{Z}_4 \times \mathbb{Z}_2$$

Group of order 8. Not cyclic. Thus not isomorphic to 1.

4.$$\mathbb{Z}_{14} \times \mathbb{Z}_5$$

Group of order 70. Abelian. Isomorphic to $\mathbb{Z}_2\times\mathbb{Z}_5\times\mathbb{Z}_7$.

5.$$\mathbb{Z}_{10} \times \mathbb{Z}_7$$

Group of order 70. Abelian. Isomorphic to $\mathbb{Z}_2\times\mathbb{Z}_5\times\mathbb{Z}_7$. Isomorphic to 4.

6.$$D_{70}$$ Symmetries of a regular 35-gon.

Group of order 70. Non-abelian. Not isomorphic to 4 or 5.

7.$$\mathcal{P}\{1,2\}$$ with symmetric difference as operation.

Group of order 4. Denote that A+A=0 for all elements A. Thus the group is isomorphic to [itex]\mathbb{Z}_2\times\mathbb{Z}_2[/tex]. Isomorphic to 2.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> 8.Fourth roots of unity </div> </div> </blockquote><br /> Group of order 8. Cyclic with generator $i^{1/2}$. Thus is isomorphic to 1.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> 9.$$\langle (12)(34),(13)(24)\rangle \leq S_4$$ </div> </div> </blockquote><br /> Group contains elements {(12)(34),(13)(24),(14)(23),1}. Group contains 4 elements and all elements have order at most 2. Isomorphic to $\mathbb{Z}_2\times\mathbb{Z}_2$. Isomorphic to 2,7.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> 10. The set generated by the matrix representations of the quaternions. </div> </div> </blockquote><br /> Group of order 8, non-abelian. Not isomorphic to 3,8<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> I don't feel confident about my answers, but I said that 2+9, 4+5, and the rest in separate classes.<br /> I grouped them first by order and then mainly examined element orders, but I'm pretty sure I left some out. Any thoughts? </div> </div> </blockquote><br /> I think the partition is: 2+7+9, 1+8, 4+5 and 3,6,10 in separate classes.[/itex]

 

[Zorba]

Thanks for the reply, but I don't think 1 can't be iso to 8 because they have different orders. Also I toyed with including 7 with 2,9, but I didn't in the end, I think you're probably right though, for some reason I thought <{1}> has order 3, damn it...

 

[micromass]

Oh, I'm sorry, the fourth roots of unity are {1,-1,i,-i}. This is indeed cyclic of order 4. Thus not isomorphic to any other group...