I am told some information about a group $G$ of order $168$. All we are told about G is that: It has one element of order one, $21$ elements of order $2$, $56$ elements of order $3$, $42$ elements of order $4$ and $48$ elements of order $7$. and later it will be proved to […]

I am going through Quantum factoring, discrete logarithms and the hidden subgroup problem by Richard Jozsa. On page 13, the author discussed the hidden subgroup problem (HSP) formulation of the graph isomorphism (GI) problem. I would like to make it sure that I get the development of the concept right. Here both $A$ and $B$ […]

This is the proof, which I mostly understand except for one bit: You have $h_1 \in H_1$ and $h_2 \in H_2$. We also have $h_1^{-1}(h_2^{-1}h_1h_2) \in H_1$, because $h_2^{-1}h_1h_2 \in h_2^{-1}H_1h_2 = H_1$. Similarly, we have $(h_1^{-1}h_2^{-1}h_1)h_2 \in H_2$. Therefore $$ h_1^{-1}h_2^{-1}h_1h_2 \in H_1 \cap H_2 = \{1_G\} $$ and so $h_1^{-1}h_2^{-1}h_1h_2 = \{1_G\}$. Let’s […]

Let $G$ be a group, $S\subset G$ a subset, then the smallest normal subgroup of $G$ which contains $S$ is called the normal closure of $S$, and denoted by $S^G$. My question is, if $G$ is a free group of rank $n$ for $n\in\mathbb N$, then is every normal subgroup of $G$ is a normal […]

$G=D_6$ and $H=<R^2>$. Use this Cayley table for $D_6$ (a). Show that $H \vartriangleleft G$. I want to show by finding out $aH=Ha$ for all $a \in G$, but then how do I proceed, it would be too tedious to check all $a$ in $D_6$, is there any way else to show it? (b). List […]

Let $O(n)$ be the standard orthogonal group of real matrices. I am trying to prove the following: $N = \bigcup_{g\in GL_n(\mathbb{R})}g\cdot O(n)\cdot g^{-1}$ is not a subgroup of $GL_n(\mathbb{R})$. I know that if it was a subgroup then it was equal to the normal closure of $O(n)$ but I do not know what that is… […]

Let $G$ group of order $pq$ where $p,q$ primes.Show that if $G$ contains normal groups $N$ and $K$ with $|N|=p$ and $|K|=q$ then is cyclic Any ideas or hints for showing this?

I recently took a course on group theory, which mentioned that the following proposition is equivalent to the continuum hypothesis: “The infinite symmetric group (i.e. the group of permutations on the set $\mathbb{N}$) has exactly 4 normal subgroups.” Does anyone have any references or explanation for this?

I am trying to prove the above with the conditions that N is normal to G and $G/N$ is abelian. So then for $Na,Nb\in{G/N}, NaNb=NbNa$. But $NaNb=Nab=Nba$, so then we know $ab=ba$, so G is abelian. So if $ab\in{G}, N=abN(ab)^{-1}=abN(ba)^{-1}=baNa^{-1}b^{-1}….$ and i’m getting lost…

Assume that $N$ is a normal subgroup of $G$ such that $N \cap G’=\{e\}$. Prove that $N \subset Z(G)$. Note 1: $Z(G)$ is the center of the group $G$, and $G’$ is the set of all commutators of $G$. Note 2: Assume that $n \in N$. We want to show that for all $g \in […]

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