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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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