0

I want to check if my solution to this problem from my abstract algebra course is correct. The problem says:

Find which of the following groups are isomorphic: $$\mathbb{Z}_8 \ \ , \ \ \mathbb{Z}_4\oplus\mathbb{Z}_2 \ \ , \ \ \mathbb{Z}_2^3 \ \ , \ \ \mathbb{Z}_{24}^\times \ \ , \ \ \mathbb{Z}_{30}^\times \ \ , \ \ D_4.$$

What I did is calculate the order of each element from each group. Doing that I obtained:

  • $\mathbb{Z}_8$ : 4 elements of order $8$, 2 elements of order $4$, 1 element of order $2$.
  • $\mathbb{Z}_4\oplus\mathbb{Z}_2$, $\ \mathbb{Z}_{30}^\times$ : 4 elements of order $4$, 3 elements of order $2$.
  • $\mathbb{Z}_2^3$, $\ \mathbb{Z}_{24}^\times$ : 7 elements of order $2$.
  • $D_4$ : 2 elements of order $4$, 5 elements of order $2$.

I don't know if this condition is sufficient to ensure that the groups that share the "number of elements of each order" are isomorphic. Am I missing something? Any help will be appreciated, thanks in advance.

Alejandro Bergasa Alonso
  • 3,639
  • 1
    A priori, there is no reason to believe that (same number of elements with same order) $\implies$ (groups are isomorphic). Why do you think that that is true? (The other direction is indeed true, which is how you can discard the possibilities of certain isomorphisms.) Are you aware of the structure theorem for finite abelian groups? – Aryaman Maithani Nov 28 '20 at 16:36
  • @AryamanMaithani my question is about what can I do next to check if the groups that share number of elements of same order are isomorphic or not. Yes, I'm aware of the structure theorem for finite abelian groups, do I need it? How do I translate the multiplicative groups so? (I struggle with multiplicative groups) – Alejandro Bergasa Alonso Nov 28 '20 at 16:39
  • 1
    For abelian groups, you can use this to see that (same number of elements with same order) is necessary and sufficient as well. For non-abelian groups, the same does not hold. Fortunately for you, only one group in your list is non-abelian and thus, you can finish it. – Aryaman Maithani Nov 28 '20 at 16:45

1 Answers1

1

Finite abelian groups are determined by the number of elements of any given order. But if you don't know the classification of abelian groups you can always be more concrete in any given case: you can map $(a,b,c) \in (\mathbb{Z}/2 \mathbb{Z})^3$ to $(-1)^a \times 5^b \times 7^c \in (\mathbb{Z}^{\times}_{24})$ and check that this is an isomorphism directly.

As noted in the comments, two groups can have the same number of elements of the same order and not be isomorphic. A good example to keep in mind is the abelian group $G = (\mathbb{Z}/3 \mathbb{Z})^3$ and the non-abelian $3$-Sylow subgroup of $\mathrm{GL}_3(\mathbb{Z}/3\mathbb{Z})$:

$$\left( \begin{matrix} 1 & * & * \\ 0 & 1 & * \\ 0 & 0 & 1 \end{matrix} \right) \subset \mathrm{GL}_3(\mathbb{Z}/3\mathbb{Z})$$

Both groups have $27$ elements and every non-trivial element has order $3$. If you don't like the description of that latter group you could think about the permutation group:

$$\langle (1, 4, 7)(2, 5, 8)(3, 6, 9),(1, 3, 2)(4, 5, 6),(4, 6, 5)(7, 8, 9) \rangle \subset S_9.$$

user855186
  • 121