# Two abelian groups are isomorphic iff their corresponding sylow subgroups are isomorphic.

I know that any abelian group can be written as a direct product of sylow subgroups and any sylow subgroup can be written as a product of cyclic groups.

Let $G$ and $G^{\prime}$ be two abelian groups of order m and n. and $m=p_1^{n_1} \dots p_k^{n_k}$ and $n=q_1^{n_1} \dots q_l^{n_l}.$ Then $G = S_1 \times \dots \times S_k,$ where $|S_i|=p_i^{n_i}$ and since any Sylow subgroups can written as a direct product of cyclic groups.

It can be written as $S_i = A_{i1} \times \dots A_{ik},$ where $A_{ij} = \langle a_{ij} \rangle,$ and $|A_{ij}| = p_i^{a_j}$ and $a_1 + \dots + a_k = n_i$ simlarly holds for $G^{\prime}.$

what can i do next ? I am struck with it ?

If $G\cong G'$, then of course also their Sylow subgroups are isomorphic. So we only need to show the converse direction. If all Sylow subgroups of $G$ and $G'$ are isomorphic, then also $G$ and $G'$ are isomorphic, since every abelian group is the direct product of its Sylow subgroups: $$G=P_1\times \cdots \times P_r\cong P_1'\times \cdots \times P_r'=G'.$$
• Because then $G$ and $G'$ can be considered "the same", and "everything" of them then is equal. – Dietrich Burde Jan 22 '18 at 19:39
• After rewriting we have $G=G'$. So of course, their Sylow subgroups are also the same, i.e., $P_i=P_i'$. What is there not to understand? Do you know what $G$ isomorphic to $G'$ means? – Dietrich Burde Jan 22 '18 at 19:41