Here is the question I am trying to understand its solution:
Prove that a group of order $11 \times 2^{10}$ has a normal subgroup.
And here is a solution I found to the part of excluding the case $n_2 = 11$ and $n_{11} = 1024.$:
Suppose that $n_2 = 11$ and $n_{11} = 1024.$ Let $\{H_i\}_{i=1}^{2^{10}}$ be the set of all $11$-Sylow subgroups of $G.$ Since $11$ is prime, if $i \neq j,$ then $H_i \cap H_j = \{e\}.$ This implies that $|\cup_{i=1}^{2^{10}}H| = 1 + 10. 2^{10}.$ Let $K_1$ and $K_2$ be $2$– Sylow subgroups of $G.$ Since $(2,11) = 1,$ this implies that $K_i \cap H_j = \{e\}.$ Since $K_1$ and $K_2$ are distinct subgroups, $|K_1 \cap K_2| > 1$ which implies $|K_1 \cup K_2| \geq 2^{10} + 1.$ This implies $|K_1 \cup K_2 \cup (\cup_{i=1}^{2^{10}} H_i)| \geq 1 + 10.2^{10} + 2^{10} = 1 + 11. 2^{10} > |G|.$ This is a contradiction.
Still I do not understand why the $2$– Sylow subgroups have a non-trivial intersection i.e., why $|K_1 \cap K_2| > 1$ while the $11$– Sylow subgroups has a trivial intersection $H_i \cap H_j = \{e\}$? could someone explain this to me, please?
Note that: I have read these questions
1 – When is the intersection between two Sylow subgroups trivial?
2- Any group of order $12$ must contain a normal Sylow subgroup
But still I do not understand the reason in my case.
Best Answer
The $11$-Sylow subgroups are groups of order $11$. If $H$ is a group of prime order, then it has no proper nontrivial subgroup; in particular, since distinct $p$-Sylow subgroups cannot contain each other, the intersection of two distinct Sylow subgroups of prime order must be trivial. So the intersection of any two $11$-Sylow subgroups, which are each of prime order, must be trivial.
That means that each element of order $11$ is in one and only one $11$-Sylow subgroup (each one is definitely in at least one, but no two distinct ones have nontrivial elements in common). How many elements do the $n_{11}=2^{10}$ subgroups account for? Each one has $10$ elements of order $11$ and one element of order $1$; the element of order $1$ is common to all of them. The rest are all distinct. So that accounts for $n_{11}(10) + 1 = 2^{10}(10)+1$ elements of the group $G$.
Now, you cannot argue the same way with the $2$-Sylow subgroups, since they aren't of prime order. But suppose first that any two $2$-Sylow subgroups have trivial intersection (that could happen). Then each of them has $2^{10}$ elements, and no non-identity element is in two or more of them. So this accounts for $n_2(2^{10}-1) + 1$ elements: $2^{10}-1$ nontrivial elements in each of the $n_2$ subgroups, plus the identity. This would give a further $11(2^{10}-1)$ elements, plus the identity (which we already counted).
So if we have both $2^{10}$ $11$-Sylow subgroups, and $11$ $2$-subgroups, and any two $2$-Sylow subgroups have trivial intersection, then this would account for $$2^{10}(11-1) + 11(2^{10}-1) + 1 = 21(2^{10}) - 10 = 11\times 2^{10} + 10(1023)\gt 11\times 2^{10}\text{ elements.}$$ As this is more than the total number of elements that we have in $G$, our assumptions cannot all hold.
The last unwarranted assumption we made was that any two of the $2$-Sylow subgroups intersect trivially. So we should assume that if $n_{11}=2^{10}$ and $n_2=11$, then there are two Sylow $2$-subgroups, call them $K_1$ and $K_2$, such that $K_1\cap K_2\neq\{e\}$. Proceed from there.