Question #289457

Let G be a group of order 112:132 . How many 11-sylow subgroups and 13 sylow subgroups are there in G?


1
Expert's answer
2022-01-23T16:53:10-0500

Let GG  be a group of order 112:132.11^2:13^2 .  

It follows from Third Sylow Theorem that for the number s11s_{11} of sylow 1111-subgroups we have that s111mod11s_{11}\equiv 1\mod 11 and s11s_{11} divides 132.13^2. Therefore, s11{1,13,169}{1,12,23,34,45,56,67,78,89,100,111,122,133,144,155,166,...}={1}.s_{11}\in\{1,13,169\}\cap\{1,12,23,34,45,56,67,78,89,100,111,122,133,144,155,166,...\}\\=\{1\}.

We conclude that s11=1,s_{11}=1, and hence there is a unique sylow 1111-subgroup in GG.


It follows from Third Sylow Theorem that for the number s13s_{13} of sylow 1313-subgroups we have that s131mod13s_{13}\equiv 1\mod 13 and s13s_{13} divides 112.11^2. Therefore, s13{1,11,121}{1,14,27,40,53,66,79,92,105,118,131,...}={1}.s_{13}\in\{1,11,121\}\cap\{1,14,27,40,53,66,79,92,105,118,131,...\}=\{1\}.

We conclude that s13=1,s_{13}=1, and hence there is a unique sylow 1313-subgroup in GG.


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