query: Structure of groups of order $pq$, where $p,q$ are distinct primes.
positive_passages.docid: 23885
positive_passages.text: Structure of a group, $G$, of order $pq$ where $p, q$ are prime.
negative_passages.docid: 2731
negative_passages.text: Groups of even order have an element whose order is 2.
negative_passages.docid: 38153
negative_passages.text: Looking for a simple proof that groups of order $2p$ are up to isomorphism $\mathbb{Z}_{2p}$ and $D_{p}$ .
negative_passages.docid: 75670
negative_passages.text: Questions around the number of subgroups of a $p$-group
negative_passages.docid: 88344
negative_passages.text: Order of a group and subsequent subgroups
negative_passages.docid: 50157
negative_passages.text: order of a sylow p-subgroup
negative_passages.docid: 49162
negative_passages.text: Classify all groups with order $pq$, $p$ is not equal to $q$
negative_passages.docid: 59811
negative_passages.text: Finite Abelian groups with the same number of elements for all orders are isomorphic
negative_passages.docid: 50706
negative_passages.text: $|G|=p(p+1)$ for $p$ prime, then $G$ has a normal subgroup of order $p$ or $p+1$
negative_passages.docid: 148904
negative_passages.text: Groups of order $p^3$
negative_passages.docid: 122970
negative_passages.text: A question on $p$-groups, and order of its commutator subgroup.
negative_passages.docid: 104080
negative_passages.text: Group with order $pq$ has subgroups of order $p$ and $q$
negative_passages.docid: 95534
negative_passages.text: If $|G|=p^nq$, then $G$ contains a unique normal subgroup of index $q$
negative_passages.docid: 121481
negative_passages.text: Non-abelian finite $p$-group in which every element has prime order?
negative_passages.docid: 28676
negative_passages.text: Is a finite group uniquely determined by the orders of its elements?
negative_passages.docid: 15630
negative_passages.text: Normal subgroup of prime order is characteristic
negative_passages.docid: 123796
negative_passages.text: Order of automorphism group of a $p$-group is divisible by $p$.