The context of this question is from the definition of the sporadic Mathieu group $M_{23}$, which (in one possible definition) is the stabilizer of a point in $M_{24}$, which is a certain subgroup of $S_{24}$ (a permutation group on $24$ points). When I read this I was a bit surprised since they haven’t specified which […]

I had asked that if $G$ is a direct product of a $2$-group and a simple group, then is it possible that $G$ be a solvable group. That the answer is no! But by a Remark on T.M. Gagen, Topics in Finite Groups, London Math. Soc. Lecture Note Ser., vol. 16, Cambridge Univ. Press, Cambridge, […]

Let $A$ be an Abelian group of prime power order. It can be expressed as a (unique) direct product of cyclic groups of prime power order: $A = \mathbb{Z}_{p^{n_1}} \times \cdots \times \mathbb{Z}_{p^{n_t}}$ where $p$ is a prime, and $n_1, \ldots, n_t $ are positive integers. Let $B$ be a subgroup of $A$. Question Is […]

I am trying to prove that any conjugate of $a$ has the same order as $a$. Let $G$ be a group and let $a \in G$. An element $b \in G$ is called a conjugate of $a$ if $b=xax^{-1}$. My professor gave us a hint. Prove $a^k=e$ iff $b^k=e$ for all $k \in \mathbb{Z}$ implies […]

This question already has an answer here: Showing there is no ring whose additive group is isomorphic to $\mathbb{Q}/\mathbb{Z}$ [closed] 1 answer

Let $W=W_{\Phi}$ be a reflection group, with root system $\Phi$, and $\Delta=\{\alpha_1, …,\alpha_n\}\subseteq \Phi$ a simple system. So $W$ is generated by the $s_{\alpha_i}=s_i$ for $i=1,2,…n. $ We know the fact that the length of the longest element $w_0$ of $W$ is $\mid\Phi^+\mid$. For example, in the type $A_3$, $w_0=s_2s_1s_3s_2s_1s_3$ is a longest element and […]

This question already has an answer here: How to prove $b=c$ if $ab=ac$ (cancellation law in groups)? 6 answers

Prove that the center of the factor group $G/Z(G)$ is the trivial subgroup $[e]$. So far I’ve proved that $Z(G)$ is a normal subgroup of $G$ for all $a \in G $.

This question already has an answer here: the image of normal subgroups 2 answers

Let $G$ be a group and $N$ a subgroup of $G$. I read that the following two definitions are equivalent: 1) $\forall g\in G$, $gNg^{-1}=N$ 2)$\forall g\in G$, $gNg^{-1}\subset N$ does this mean that we have always $N\subset gNg^{-1}$ ? my guess is no because take $n\in N$ then if $n=gn'g^{-1}$ then $n'=g^{-1}ng$ and there […]

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