This is what I have so far:

$$\mathbb{Z}_{12} = \{0, 1, 2,\cdots, 11\}$$ $$\mathbb{Z}_{18} = \{0, 1, 2,\cdots, 17\}$$ $$\mathbb{Z}_9 = \{0, 1, 2,\cdots, 8\}$$ $$\mathbb{Z}_4 = \{0, 1, 2, 3\}$$

Using that fact that isomorphisms must contain the same number of elements, we must find a subgroup in $\mathbb{Z}_{12}\oplus \mathbb{Z}_{18}$ with the same number of elements as in $\mathbb{Z}_{9}\oplus \mathbb{Z}_{4}$.

Now, let's determine the number of elements in $\mathbb{Z}_{9}\oplus \mathbb{Z}_{4}$: $$|\mathbb{Z}_{9}\oplus \mathbb{Z}_{4}| = \text{lcm}(|\mathbb{Z}_{9}|, |\mathbb{Z}_{4}|) = \text{lcm}(|9|, |4|) = 36$$

Next, let's find a subgroup in $\mathbb{Z}_{12}\oplus \mathbb{Z}_{18}$ with order $36$. First, $\langle 3 \rangle$ in $\mathbb{Z}_{12}$ gives us the group $\{0, 3, 6, 9\}$ which has order $4$. Second, $\langle 2\rangle$ in $\mathbb{Z}_{18}$ gives us the group $\{0, 2, 4, 6, 8, 10, 12, 14, 16\}$ which has order $9$. Then, the group generated by $(3, 2)$ has order $36$ (because $\text{lcm}(9,4) = 36$)

My question is, does this properly show that there exists a subgroup in $\mathbb{Z}_{12}\oplus \mathbb{Z}_{18}$ that is isomorphic to $\mathbb{Z}_{9}\oplus \mathbb{Z}_{4}$? Or do I have to show that the mapping is one-to-one, onto, and preserves the group operation? If so, what is the mapping function to show one-to-one, onto, and operation preservation?


  • 3
    $\begingroup$ Order is not enough: Note that $\mathbb{Z}_2+\mathbb{Z}_2$ and $\mathbb{Z}_4$ have the same order, but are not isomorphic. To use your approach, I would say that $\mathbb{Z}_{12}$ has a subgroup isomorphic to $\mathbb{Z}_4$ (you produced it) while $\mathbb{Z}_{18}$ has a subgroup isomorphic to $\mathbb{Z}_9$ (you produced it). Therefore $\dots$. Whether exhibiting the isomorphisms should be done is course-dependent. Another approach is to note that $\mathbb{Z}_{12}$ is isomorphic to the direct sum $\mathbb{Z}_4+\mathbb{Z}_3$, while the other is isomorphic to the direct sum $\dots$, so $\dots$. $\endgroup$ – André Nicolas Nov 12 '13 at 1:23
  • $\begingroup$ The equation you wrote $|\mathbb{Z}_9 \oplus \mathbb{Z}_4| = \operatorname{lcm}(9,4)$ is technically true here (9 and 4 are coprime), but the order of a direct sum is always the product, not the lcm. For example the order of $|\mathbb{Z}_2 \oplus \mathbb{Z}_2|$ is 4, not $\operatorname{lcm}(2,2)=2$. $\endgroup$ – Najib Idrissi Nov 12 '13 at 1:41

Hint: Is there an element in $\mathbb{Z}_{12}$ that doesn't generate all of $\mathbb{Z}_{12}$ (hence is a factor of $12$) but generates $\mathbb{Z}_{9}$ or $\mathbb{Z}_4$? Then is there an element in $\mathbb{Z}_{18}$ not relatively prime to $18$ that generates $\mathbb{Z}_9$ or $\mathbb{Z}_4$?

Call these two elements $a$ and $b$. Then look at the element $a \oplus b \in \mathbb{Z}_{12} \oplus \mathbb{Z}_{18}$. Create an 'obvious' mapping from this element into $\mathbb{Z}_9\oplus \mathbb{Z}_4$. Prove that this mapping must be $1:1$, onto, and preserves the group operation.

But makes things easy on yourself! Focus on the element $a \oplus b$, it should generate $\mathbb{Z}_9 \oplus \mathbb{Z}_4$ (why?) then those three things are easy to prove. (again, think about why).

EDIT: I'll expand a bit more on my hint then. You already looked at a few elements in $\mathbb{Z}_{12}$. Look closely at the subgroup $\langle 3 \rangle=\{0,3,6,9\}$. Could we map that to $\mathbb{Z}_4$ with a mapping that preserves the operation, say $\varphi$? Now look at $\langle 2 \rangle =\{0,2,4,6,8,10,12,14,16\}\subset \mathbb{Z}_{18}$. Can we take those elements and map it to $\mathbb{Z}_9$, say the mapping is $\theta$, in a way that preserves the operation? Now since clearly $\mathbb{Z}_{12} \oplus \mathbb{Z}_{18} \cong \mathbb{Z}_{18} \oplus \mathbb{Z}_{12}$, would the 'mapping'

$$\psi((a,b))=\varphi(a) \oplus \theta(b)$$

for $(a,b) \in \mathbb{Z}_{18} \oplus \mathbb{Z}_{12}$ be an isomorphism to $\mathbb{Z}_{18} \oplus \mathbb{Z}_{12}$?

FINAL EDIT: With your last comment you essentially have the idea. It is clear that $\mathbb{Z}_{12} \oplus \mathbb{Z}_{18} \cong \mathbb{Z}_{18} \oplus \mathbb{Z}_{12}$ (just take the map that sends $(a,b)$ to $(b,a)$). So we just need to find a subgroup of $\mathbb{Z}_{18} \oplus \mathbb{Z}_{12}$ isomorphic to $\mathbb{Z}_9 \oplus \mathbb{Z}_4$. Your idea is right. We make a mapping $\varphi$ that takes each element in $\mathbb{Z}_{18}$ mod $2$ and a mapping $\theta$ that takes each element in $\mathbb{Z}_{12}$ mod $3$. Then show that $\varphi \oplus \theta$ is an isomorphism. Now here is where you need to be careful. You said it would be an isomorphism of $\mathbb{Z}_4 \oplus \mathbb{Z}_9$ to $\mathbb{Z}_{18} \oplus \mathbb{Z}_12$. This is wrong. It's an isomorphism from $\mathbb{Z}_{4} \oplus \mathbb{Z}_9$ to $\langle 2 \rangle \oplus \langle 3 \rangle$ in $\mathbb{Z}_{18} \oplus \mathbb{Z}_{12}$ (so a subgroup of that group). Since this group is isomorphic to $\mathbb{Z}_{12} \oplus \mathbb{Z}_{18}$, we would be done since it would be clear what the subgroup was. But there is an easier way to do this.

Let's look at the subgroup $\langle 2 \rangle \oplus \langle 3 \rangle$ of $\mathbb{Z}_{18} \oplus \mathbb{Z}_{12}$. Let's create a mapping $\varphi$ of $\langle 2 \rangle$ in $\mathbb{Z}_{18}$ into $\mathbb{Z}_9$. Notice that $1$ generates $\mathbb{Z}_9$. Moreover, $\langle 2 \rangle$ generates the subgroup (itself) in $\mathbb{Z}_{18}$. So just map $2 \in \mathbb{Z}_{18}$ to $1 \in \mathbb{Z}_{9}$. Similarly, we can map $3 \in \mathbb{Z}_{12}$ to $1 \in \mathbb{Z}_4$. Since we are mapping generators to generators, it is clear that we preserve the operations, it is $1:1$, and onto. So we have the required isomorphism. But all the same, if this shorter way isn't as clear to you then you can can always show it by the preceding paragraph (which was what you stated in your comment, I just cleaned up a bit of the language). Hope that helped finally clear things up.

  • $\begingroup$ I guess I'm confused at your hint. I did not find an element in Z12 that "doesn't generate all of Z12 but generates Z9 or Z4". The potential candidates are 1, 2, 3, 4, and 6 because all are factors of 12. <1> = Z12, <2> = {0, 2, 4, 6, 8, 10}, <3> = {0, 3, 6, 9}, <4> = {0, 4, 8}, <6> = {0, 6}. $\endgroup$ – JT9 Nov 12 '13 at 4:29
  • $\begingroup$ @JTWheeler Yes, now count the amount of elements in those subgroups. Map the generators from one of those subgroups of the appropriate size to the $\mathbb{Z}_{-}$ you want. Do it for both then 'stitch' the maps together and show they typical 3 things you need for an isomorphism. I have elaborated more in my answer above to help with this. Focus on the generators! $\endgroup$ – mathematics2x2life Nov 12 '13 at 6:18
  • $\begingroup$ I think I see what you're getting at now. <3> in Z12 can be mapped to Z4 by dividing each element by 3. So the mapping function would be a/3 to achieve Z4. <2> in Z18 can be mapped to Z9 by diving each element by 4. So the mapping function would be b/4 to achieve Z9. I believe I now have to show 1-1, onto, and operation preservation for these 2 functions. And then finally show the mapping ψ((a,b))=φ(a)⊕θ(b) is an isomorphism to ℤ18⊕ℤ12. $\endgroup$ – JT9 Nov 12 '13 at 16:31
  • $\begingroup$ @JTWheeler Not quite but you essentially have the exact idea. I'll do one final edit to explain what you just said in a bit rigor. $\endgroup$ – mathematics2x2life Nov 12 '13 at 16:39
  • $\begingroup$ I think the second part of your final edit makes more sense to me actually. One final question for you: ℤ12⊕ℤ12≅ℤ18⊕ℤ12 (I think you meant ℤ12⊕ℤ18≅ℤ18⊕ℤ12), is this a property of external direct products? $\endgroup$ – JT9 Nov 12 '13 at 17:21

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