Let G be a group of order 112:132.
It follows from Third Sylow Theorem that for the number s11 of sylow 11-subgroups we have that s11≡1mod11 and s11 divides 132. Therefore, s11∈{1,13,169}∩{1,12,23,34,45,56,67,78,89,100,111,122,133,144,155,166,...}={1}.
We conclude that s11=1, and hence there is a unique sylow 11-subgroup in G.
It follows from Third Sylow Theorem that for the number s13 of sylow 13-subgroups we have that s13≡1mod13 and s13 divides 112. Therefore, s13∈{1,11,121}∩{1,14,27,40,53,66,79,92,105,118,131,...}={1}.
We conclude that s13=1, and hence there is a unique sylow 13-subgroup in G.
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