# Non-isomorphic abelian groups of order 8

$\newcommand{\Z}{\mathbb Z}$

I don't understand why the following groups are not isomorphic with each other $\Z/8\Z,$ $\Z/4\Z \times \Z/2\Z,$ and $(\Z/2\Z)^3$.

Indeed, i thought because they re all abelian finite groups with order equals $8$. So they were isomorphic.

thanks

• If you look at groups of order $4$, both $\Bbb{Z}/4\Bbb{Z}$ and $\Bbb{Z}/2\Bbb{Z} \times \Bbb{Z}/2\Bbb{Z}$ are abelian groups of order $4$, but are they isomorphic? – ÍgjøgnumMeg Jun 18 '17 at 23:07
• But you're wrong $\mathbb Z_8 \not\cong \mathbb Z_4 \times \mathbb Z_2$, etc. $\mathbb Z_m\times \mathbb Z_n = \mathbb Z_{m,n} \iff \gcd(m, n) = 1$. Similarly .... – Namaste Jun 18 '17 at 23:11
• thanks for answering me So why, according to the theorem of structure of finite-type abelian groups, to find all the abelian groups of order 8 it suffices to determine all the possibilities for the elementary divisors of Z / 8Z? I do not understand the difference, what is the criterion for two Abelian groups of the same cardinal to be isomorphic? – Farouk Deutsch Jun 18 '17 at 23:12
• See the last part of my comment above: note that $\gcd(4, 2) = \gcd(2, 2, 2) = 2 \neq 1$. On the other hand, $\mathbb Z_{15} \cong \mathbb Z_3\times \mathbb Z_5$ because $\gcd(3, 5) = 1.$ – Namaste Jun 18 '17 at 23:15

The group $\mathbb{Z}/8\mathbb{Z}$ has an element of order $8$.
The group $(\mathbb{Z}/4\mathbb{Z})\times(\mathbb{Z}/2\mathbb{Z})$ has an element of order $4$ but no element of order $8$.
The group $(\mathbb{Z}/2\mathbb{Z})^3$ has no element whose order is greater than $2$.
• the fundamental theorem says: any finite abelian group is isomorphic to something like this $\mathbb{Z}/d_1\mathbb{Z} \oplus \dots \oplus \mathbb{Z}/d_k\mathbb{Z}$ where $d_{i+1} | d_i$. So any two finite abelian groups would be isomorphic if when you put them in such a form, they were the same. – mdave16 Jun 18 '17 at 23:17
• @FaroukDeutsch The thorem states (for finite groups) that the group can be obtained as $(\mathbb{Z}/n_1\mathbb{Z})\times(\mathbb{Z}/n_2\mathbb{Z})\times\cdots(\mathbb{Z}/n_k\mathbb{Z})$, with $n_1\mid n_2$, $n_2\mid n_3$ and so on. In the case of $8$, you have three possibilities: $8$, $2$ and $4$, and $2$ three times. – José Carlos Santos Jun 18 '17 at 23:18