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.

I am reading in my book about Cyclic groups and ran into this example that I would like a little more explanation on why or why not each group is isomorphic.

a) $\mathbb{Z}_2 \times \mathbb{Z}_2$ and $\mathbb{Z}_4$

b) $\mathbb{Z}^*_{12}$ and $\mathbb{Z}^*_8$

c) $\mathbb{Z}^*_5$ and $\mathbb{Z}_4$

d) $\mathbb{Z}_2 \times \mathbb{Z}$ and $\mathbb{Z}$

e) $\mathbb{Q}$ and $\mathbb{Z}$

f) $\mathbb{Z} \times \mathbb{Z}$ and $\mathbb{Z}$

Thanks in advance.

share|improve this question

closed as not constructive by Carl Mummert, Davide Giraudo, user1729, Julian Kuelshammer, Arkamis May 1 '13 at 13:28

As it currently stands, this question is not a good fit for our Q&A format. We expect answers to be supported by facts, references, or expertise, but this question will likely solicit debate, arguments, polling, or extended discussion. If you feel that this question can be improved and possibly reopened, visit the help center for guidance.If this question can be reworded to fit the rules in the help center, please edit the question.

1  
Do you have any ideas on which are/are not isomorphic? I think it's valuable for a student to develop his/her own intuition about this stuff before cranking out a formal proof. –  Christopher A. Wong Apr 1 '13 at 22:02
    
Yeah, I understand the general concepts such that to have a group be isomorphic it needs to have a group isomorphism exist where its group homomorphism is bijective. –  Mike Apr 1 '13 at 22:13
    
@Mike: The reason, I believe, that this thread has been closed is because people would like to know what you have actually tried. I think it would be helpful if, for example, you edited your question to give us some of your thoughts on question (a). Do you think they are isomorphic or not? (Hint: $\mathbb{Z}_4$ is cyclic. So every isomorphic group is cyclic (why?)...etc.) –  user1729 Apr 2 '13 at 13:14
3  
I don't think 7 downvotes are appropriate and hence I upvoted to compensate. –  Matt N. Apr 4 '13 at 7:31
1  
Hint: if you show what you've tried or what is giving you trouble, the question might get more upvotes (or fewer downvotes). This also helps those answering to focus their efforts. –  robjohn Apr 4 '13 at 10:36
show 5 more comments

2 Answers 2

Why not draw yourself some group tables? In the case of $\mathbb{Z}_2 \oplus \mathbb{Z}_2$ we have $\mathbb{Z}_2 \oplus \mathbb{Z}_2 = \{(0,0),(0,1),(1,0),(1,1)\}$

\begin{array}{c|cccc} +& (0,0) & (1,0) & (0,1) & (1,1) \\ \hline (0,0) & (0,0) & (1,0) & (0,1) & (1,1) & \\ (1,0) & (1,0) & (0,0)& (1,1) & (0,1) & \\ (0,1) & (0,1) & (1,1) & (0,0) & (1,0) & \\ (1,1) & (1,1) & (0,1) & (1,0) & (0,0)& \\ \end{array}

Next, look at the group $\mathbb{Z}_4$.

\begin{array}{c|cccc} +& 0 & 1 & 2 & 3 \\ \hline 0 & 0& 1& 2 & 3 & \\ 1 & 1& 2& 3 & 0& \\ 2 & 2& 3& 0 & 1 & \\ 3 & 3& 0& 1 & 2& \end{array}

Can you see a bijection between $\mathbb{Z}_2 \oplus \mathbb{Z}_2$ and $\mathbb{Z}_4$ that preserves the group structure?

In other cases, you can draw tables, or you can ask yourself simple questions like "Do they have the same number of elements?" For example, a finite group can't be isomorphic to an infinite group.

share|improve this answer
    
Notice that $\mathbb{Z}_2\oplus\mathbb{Z}_2$ has characteristic $2$. That is, for each $x\in\mathbb{Z}_2\oplus\mathbb{Z}_2$, $x+x=0$. Consider $1\in\mathbb{Z}_4$; $1+1=2\ne0$. –  robjohn Apr 4 '13 at 10:34
1  
@robjohn That is what I wanted the OP to establish for themselves. –  Fly by Night Apr 4 '13 at 12:28
    
Yes, but since they hadn't responded, I thought a further hint might be useful. –  robjohn Apr 4 '13 at 12:33
add comment

To add to Fly by Night's answer: Consider $\mathbb Z \times \mathbb Z$ and $\mathbb Z$ and try to define an isomorphism $\varphi$. Assuming $+$ is defined as $(a,b) + (c,d) = (a + c, b + d)$ on $\mathbb Z \times \mathbb Z$. Note that $\mathbb Z$ is generated by $1$ hence is cyclic. Assume $\mathbb Z \times \mathbb Z$ was generated by an element $(a,b)$. Then from $n(a,b) = (0,1)$ and $m (a,b) = (1,0)$ you deduce that both $a$ and $b$ must be zero. Since $(0,0)$ is the neutral element in $\mathbb Z \times \mathbb Z$, it certainly does not generate it. We have therefore shown that $\mathbb Z \times \mathbb Z$ cannot be cyclic.

Since cyclicity is preserved by group isomorphisms, $\mathbb Z \times \mathbb Z$ and $\mathbb Z$ cannot be isomorphic.

share|improve this answer
add comment

Not the answer you're looking for? Browse other questions tagged or ask your own question.