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Jul
15
revised Relatively prime to $42$ and $70$
deleted 1 character in body
Jul
15
answered Relatively prime to $42$ and $70$
Jul
15
comment Let $a,b \in$ group $G$ such that $ab^3a^{-1}=b^2, b^{-1}a^2b=a^3$ Prove that $a=b=e$ (identity)
In the finite case: the relations show that both $a^2$ and $a^3$ are conjugate, whence have the same order. This can only happen if $a=1$. Same reasoning for $b$.
Jul
13
comment Showing Sylow $p$-groups are Abelian if $n_p=2p+1$
You wrote, this is a qual exam, study Theorem 1.1 of Isaacs' Finite Group Theory book for example. You will find it on Googel Books.
Jul
13
comment Showing Sylow $p$-groups are Abelian if $n_p=2p+1$
$G$ acts on the left cosets of the normalizer by left multiplication (what I am using here is sometimes referred to as the $n!$ theorem). The kernel of this action is trivial, since $G$ is simple.
Jul
13
revised Showing Sylow $p$-groups are Abelian if $n_p=2p+1$
added 57 characters in body
Jul
13
answered Showing Sylow $p$-groups are Abelian if $n_p=2p+1$
Jul
12
revised Prove a group is not simple, and nilpotent
edited title
Jul
12
answered Prove a group is not simple, and nilpotent
Jul
11
revised Regarding subspace of Hilbert space
edited title
Jul
11
comment Regarding subspace of Hilbert space
@user72012 you mean when $x \neq 0 \neq y$.
Jul
11
answered Finding argument of complex number and conversion into polar form
Jul
11
revised Let G be a group, $N<K<G$ and $N\trianglelefteq G$. Prove that $K/N \trianglelefteq G/N$
added 167 characters in body
Jul
11
answered Let G be a group, $N<K<G$ and $N\trianglelefteq G$. Prove that $K/N \trianglelefteq G/N$
Jul
10
comment Suppose $ G = G_{1}G_{2} $. For any prime $ p $, there exist $ P , P_{1} , P_{2} $ such that $ P = P_{1}P_{2} $
Put $G_1=H$ and $G_2=K$. I simplified your (Huppert's) notation, to get rid of all the subindices.
Jul
10
comment Suppose $ G = G_{1}G_{2} $. For any prime $ p $, there exist $ P , P_{1} , P_{2} $ such that $ P = P_{1}P_{2} $
This IS the theorem!
Jul
10
comment Let $G$ be a group of finite order, $H$ and $K$ subgroups so that $H \unlhd G$; $K \unlhd HK \unlhd G$ and $(|H|,|K|) = 1$. Show that $K \unlhd G$.
Yes of course. I will not spell it out. Let $g \in G$, then since $HK \unlhd G$ we have $g^{-1}HKg=HK$. So $g$ induces an automorphism of $HK$. Now under this automorphism $K$ is sent to $g^{-1}Kg:=L$. $L$ is again a normal subgroup of $HK$. Look at $KL$ and try to deduce from the isomorphism theorem and the fact that $K$ and $L$ have the same order, relatively prime with their index, that $K=L$.
Jul
9
answered Product of all elements in finite group
Jul
9
revised Let $G$ be a group of finite order, $H$ and $K$ subgroups so that $H \unlhd G$; $K \unlhd HK \unlhd G$ and $(|H|,|K|) = 1$. Show that $K \unlhd G$.
deleted 3 characters in body
Jul
9
answered Let $G$ be a group of finite order, $H$ and $K$ subgroups so that $H \unlhd G$; $K \unlhd HK \unlhd G$ and $(|H|,|K|) = 1$. Show that $K \unlhd G$.