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8

A $p$-group is a group where the order of every element is a power of $p$. A direct consequence of this is that if the order of every element is a power of $p$, then the order of the group must be a power of $p$. This follows from Lagrange's theorem: the order of a subgroup of a group divides the order of the group. So, one can say that a $p$-group is a ...


6

$2014 = 2 * 19 * 53$ As $2, 19, 53$ are all distinct, there is only one Abelian group of order $2014$ up to isomorphism, which is $\mathbb{Z}_{2014}\simeq\mathbb{Z}_{2}\oplus\mathbb{Z}_{19}\oplus\mathbb{Z}_{53}$. Also you can proceed this way without using the classification theorem. By Cauchy's theorem there are elements of $2, 19, 53$. Suppose they are ...


2

Yes, you can have nontrivial homomorphisms. For instance, from $$(\mathbb Z, +)$$ to $$ (\mathbb Z/3\mathbb Z, +) $$ you can send $n$ to $n \bmod 3$. Consider the matrix $$ A = \begin{bmatrix} 0 & 1 & 0\\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix} $$ Since $A^3 = I$, the map that sends $n \in (\mathbb Z, +)$ to $A^n$ in $P(3; \mathbb ...


2

Unfortunately, no. In fact, it's even worse! The quaternion group $Q_8$ has subgroups of all possible orders, and every subgroup is normal, but the group isn't abelian! More generally $p$-groups, groups whose orders are a power of a prime, will have subgroups of all orders, yet need not be abelian. There is a partial converse to Lagrange's theorem, about ...


2

Yes. It follows easily from the fact that maximal subgroups of finite supersolvable groups have prime index.


1

Yes: let $K,L,M$ the three proper subgroups of $V_4$, then $G=\pi^{-1}(K)\cup\pi^{-1}(L)\cup\pi^{-1}(M)$ (where $\pi$ is the quotient projection $x\mapsto x+H$) because $V_4=K\cup L\cup M$ and the counterimage of a proper subgroup under a surjective homomorphism is proper. (In fact $\pi(\pi^{-1}(K))=K\cap\pi(G)=K$)


1

Hint: Recall that if $H$ and $K$ are finite subgroups of $G$, then $$o(HK)=\frac{o(H)o(K)}{o(H\cap K)}$$ I am using $o(L)$ to denote the order of $L$. You also know $HK$ may itself not be a subgroup, but certainly every element of $HK$ is an element of $G$.


1

I will call $T$-groups the groups $G$ such that whenever $A_d$ is non-empty, $Aut(G)$ acts transitively on $A_d$. We begin with the $p$-group case, then nilpotent case then... $\textbf{1}$ Suppose $G$ is an abelian $T$-group (it is also a $p$-group) then $G$ is isomorphic to $\mathbb{Z}_p^n\text{ or }\mathbb{Z}_{p^n} $. The reason for this is ...



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