Let G be a finite group with&nbsp;card(G)=p2q&nbsp;with&nbsp;p&lt;q&nbsp;two ' numbers. We denote&nbsp;sq&nbsp;the number of q-Sylow subgroups of G and similarly for p. I have just shown that&nbsp;sq∈{1,&nbsp;p2}. Now I want to show that∪S∈Sylq(G)S∖{1}=∪˙S∈Sylq(G)S∖{1}i.e. that for&nbsp;S,&nbsp;T∈Sylq(G)&nbsp;with&nbsp;S≠T&nbsp;we that&nbsp;S∖{1}∩T∖{1}=∅

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Let G be a finite group with&amp;nbsp;card(G)=p2q&amp;nbsp;with&amp;nbsp;p&amp;lt;q&amp;nbsp;two &#039; numbers. We denote&amp;nbsp;sq&amp;nbsp;the number of q-Sylow subgroups of G and similarly for p. I have just shown that&amp;nbsp;sq∈{1,&amp;nbsp;p2}. Now I want to show that∪S∈Sylq(G)S∖{1}=∪˙S∈Sylq(G)S∖{1}i.e. that for&amp;nbsp;S,&amp;nbsp;T∈Sylq(G)&amp;nbsp;with&amp;nbsp;S≠T&amp;nbsp;we that&amp;nbsp;S∖{1}∩T∖{1}=∅

Drantumcem0 · Accepted Answer

Step 1S∩Tis a subgroup of S and T, then its order divides|S|=|T|=qSince q is a &#039;, then either |S∩T|=1 or |S∩T|=q.Since S≠T it must hold |S∩T|=1, which means S∩T={1G}, therefore (S∖{1G})∩(T∖{1G})=∅

Let G be a finite group with

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