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