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8 votes
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Center of group of order 3773

Let $P$ be a $7$-Sylow subgroup, $Q$ is an $11$-Sylow subgroup. Then $P,Q$ are normal in $G$, $P\cap Q=\{e\}$ and by comparing cardinalities, $G=PQ$. It is well known that in this case we have: $G\...
Mark's user avatar
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6 votes
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When does every element of Sylow $p$-subgroup is also a member of another Sylow $p$-subgroup?

Here's an example of order $180$. The idea is to find a group with a Sylow subgroup $P$, such that for every element $x\in P$, we have \begin{equation*} C_G(x)\not\le N_G(P) \end{equation*} This is ...
Steve D's user avatar
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5 votes
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G has a element of order 2 not lying in center

Let $g\in G$ be the element of order two not lying in the center. Let $G$ act by conjugation on the set of left cosets of $G/\langle g\rangle .$ So you get a homomorphism from $G$ into $S_4.$ The ...
calc ll's user avatar
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4 votes

Center of group of order 3773

If $G$ is abelian, then you are done. So, let's suppose $G$ nonabelian. From $n_{11}=1$ follows that the Sylow $11$-subgroup is normal and, as such, it is union of conjugacy classes. Now, the ...
Kan't's user avatar
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3 votes
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Prove the index of a proper subgroup of a simple group of order 17971200 is at least 14.

Let $G$ be a nontrivial finite group and $X_G$ the set of all the proper subgroups of $G$, $X_G:=\{H\le G\mid H\ne G\}$. Lagrange's theorem already puts a limitation on the order of the elements of $...
Kan't's user avatar
  • 3,137
2 votes
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Find a group of order 5784 that does not have a normal subgroup of index 12

Posting the answer I arrived at with the kind help of @SteveD in the comment thread above: Consider the group 𝐺 = $S_4 \times C_{241}$. As the direct product of $S_4$ (the symmetric group on 4 ...
giorgio's user avatar
  • 423
2 votes

Classify, up to isomorphism, all groups of order $5\cdot 7 \cdot 19$.

This is a particular case of the general setting: $|G|=pqr$, $p<q<r$, $p\nmid q-1$, $p\nmid r-1$ and $q\nmid r-1$. By Sylow III, $n_p\mid qr$ and $n_p\equiv 1\pmod p$. So, $n_p=1,q,r,qr$ and $...
Kan't's user avatar
  • 3,137
2 votes
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Do all the conjugates of $H\le G$ pairwise intersect in subgroups of the same order?

For the general case, let $G = \mathbb{Z}/2\mathbb{Z} \wr \mathbb{Z}/4\mathbb{Z} = (\mathbb{Z}/2\mathbb{Z})^{\{1, 2, 3, 4\}} \rtimes \mathbb{Z}/4\mathbb{Z}$. Then $H_1 = (\mathbb{Z}/2\mathbb{Z})^{\{1, ...
David Gao's user avatar
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1 vote
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Non-Abelian group of order $21$ subgroups

I) Your answer is not correct. 21 is ruled out because G is nonabelian, whereas subgroups of order 3 and 7 exist by Cauchy's theorem (or, more elementarily, because the nontrivial elements must have ...
Kan't's user avatar
  • 3,137
1 vote

Group of order $p^{\alpha}q$ is not simple.

Since the Sylow $p$-subgroups are all pairwise conjugate (Sylow II), their intersection is the normal core in $G$ of any of them. So, once you know this intersection is not trivial, you get a ...
Kan't's user avatar
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