Hot answers tagged

5

This is not true in general., For example $\Bbb Z/4\Bbb Z$ is not isomorphic to $\Bbb Z/2\Bbb Z\times \Bbb Z/2\Bbb Z$. You probably also need to assume both groups are cyclic. In that case you can write one as $\{a,a^2,\dots,a^n\}$ and the other as $\{b,b^2,\dots,b^n\}$. Then map $a\mapsto b$ to get the isomorphism.


3

$\Bbb Z_6$ is commutative and $S_3$ isn't, and both have six elements.


3

You say $X$ is a finite set, not a group. Thus "on group $X$" (quoting your question as it now stands) is wrong. One may say $G$ is a transitive permutation group on the set $X$, but not on the group $X$ unless $X$ is a group. It makes sense to speak of a transitive permutation group on a set $X$, but one does not say "$G$ is a transitive permutation". ...


3

Hint. Your aim is wrong: you don't need to show that $ab$ commutes with $a'b'$, you need to show that $a$ commutes with $b$. So, $$\mu((a,b)(a,b))=\cdots\,?$$


2

No, it doesn't work since this result is wrong. For example $\mathbb Z/4\mathbb Z$ is not isomorphic to $\mathbb Z/2\mathbb Z\times \mathbb Z/2\mathbb Z$. To prove that there are not isomorphic, simply remark that $\mathbb Z/4\mathbb Z$ has an element of order $4$ whereas $\mathbb Z/2\mathbb Z\times \mathbb Z/2\mathbb Z$ has no element of order $4$.


2

Let $x \in Q$. The order of $h(x)$ divides the order of $x$, hence divides $q^m$. But $h(x) \in Aut(C_p) \cong C_{p-1}$, hence the order of $h(x)$ also divides $p-1$. If $q^m$ and $p-1$ are co-prime, we deduce that the order of $h(x)$ is $1$ for any $x$. This means that $h$ is trivial, so that the semi-direct product is actually a direct product.


1

By Cauchy's theorem, there must be an element $s$ of order $2$ in $G$, and an element $r$ of order $3$, because $2$ and $3$ divide $6$. Clearly $r$ generates a cyclic group $C_3$ which has index $2$ in $G$, hence is a normal subgroup. Then $G$ is a semidirect product of $C_2$ and $C_3$. In case it is abelian, this product is direct. One can say this more ...


1

The fact that $HN=G$ tells you that, if $g\in G$, then $g=xy$ for some $x\in H$ and $y\in N$. Then $$ gN=xyN=xN $$ because $yN=N$. What precisely is the first isomorphism theorem is a bit vague: the terminology is imprecise and varies from author to author. In any case a basic theorem in group theory is that, under the assumption that $N$ is a normal ...



Only top voted, non community-wiki answers of a minimum length are eligible