So I’ve been reading this wonderful PDF textbook on algebraic topology: http://www.math.uchicago.edu/~may/CONCISE/ConciseRevised.pdf In particular, I’m very interested in the chapter on graphs. This corollary seems to give the construction of the fundamental group of a graph in fairly simple terms (here $\chi = V – E$): Corollary. If X is a connected graph, then $\pi_1(X)$ […]

Problem Suppose a finite set $G$ is closed under an associative product and that both cancellation law hold in $G$. Prove that $G$ must be a group. Also show by an example that if one just assumed one of the cancellation laws holds, then the conclusion need not follow. This is from Topics in Algebra […]

If $G$ and $H$ are divisible groups each of which is isomorphic to a subgroup of the other, then $G$ is isomorphic to $H$. Here, $G$ and $H$ are abelian groups. Can we assume another adjective rather than divisibility?

How do you prove a group of order 27 can be non abelian? From what I tried: Say $|G|=27$ then it’s a p-group therefore the ceneter is non trivial so $|Z(G)|$ could be either $3,9,27$ if $|Z(G)|=27$ then the group is abelian. if $|Z(G)|=9$ then from lagrange $|G/Z(G)|=3$ therefore cyclic therefore $G$ is abelian (which […]

Show that group of all real matrices of form $$ \begin{bmatrix} x & y\\ -y & x \end{bmatrix} , \qquad (x,y) \ne (0,0) $$ is isomorphic with/to $\mathbb C \setminus \left\{{0}\right\}$ under complex multiplication? I know two ways to show isomorphism: 1) finding a homomorphic function 2) writing the multiplication table and comparing. I think […]

Suppose $G$ is a non-abelian group and $H,K$ are two abelian subgroups of $G$. Then must $HK$ be an abelian subgroup of $G$? I know an example, but I am confused. Thus I just want to check that.

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 […]

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