# Tag Info

10

Possibly the smallest group where this occurs is the alternating group $A_{4},$ which has order $12$ and yet has no subgroup of order $6$. Note that this is a solvable group.

5

It is not possible for $G$ to be finite and abelian for this to be true. Many examples are given with $n=2m$ by simple groups: any subgroup of index 2 is normal, hence no simple group can have a subgroup of index 2. Furthermore, any nonabelian simple group is of even order, so any nonabelian finite simple group gives an example. As Hagen von Eitzen points ...

3

In the finitely generated case the answer is NO as pointed out by @Timbuc in the comments. In the infinitely generated case I believe that the answer is YES by the following construction. Let $K \subseteq \mathbb{N}$ be a recursively enumerable non-recursive set and consider a countable abelian group given by the following presentation (commutators ...

3

Hints: Fix decompositions for $A,B,C$ and note that they "stack" to become decompositions of $A\oplus B$ and $A\oplus C$. The components of the decompositions must line up exactly, and without loss of generality, we can line up the components of $A$ on both sides because they're identical. What does this say about the remaining components?

3

For your first question: If the proposition just said $n$ coprime integers, that would (or could) be taken to mean that each pair of integers $\{v_i,v_j\}|i \neq j$ is coprime ($\gcd(v_i,v_j) = 1$). This is a much stronger condition than is required for the proposition (which is that there is no "global" divisor $d>1|d\in\Bbb{Z}$ dividing every $v_i$. ...

3

Indeed it is not cyclic: since $\sqrt{2}$ is irrational, the only way the sum is an integer is if $b=0$, so this means there are no additive relations between $a$ and $b$ and the group is isomorphic to it is isomorphic to $\Bbb Z\oplus\Bbb Z$.

3

$$\mathbb{Z}_m\oplus \mathbb{Z}_n=\mathbb{Z}_{mn}.$$ And from here it is easy to show that all the subgroups are $$\mathbb{Z}_k,$$ where $k$ is divisor of $mn$.

2

If you stick to the "$QAP^{-1}$ convention" and do things horizontally, then $(Q^{-1})^{\top}$ is what gives you the basis and $P^{-1})^{\top}$ is what gives you the generators. To avoid transposing, take $\begin{pmatrix}1&1&4\\2&-1&-2\end{pmatrix}$. Using elementary row and column operations and recording them in matrix form, I get that ...

2

This is never true for non-abelian groups if $n=2$, but otherwise, for $n\ge 3$ this can happen for certain non-abelian groups. For more details see here. Example: The quaternion group satisfies $(ab)^n=a^nb^n$ for all $n=2^k$ with $k>1$, but is not abelian.

2

$n\mathbb{Z}_p\subseteq G$ is just Lagrange's theorem applied to the quotient group $\mathbb{Z}_p/G$ in the form: $|\mathcal G|=n, x\in \mathcal G \implies x^n =1$.

2

Let $\{P\}=Syl_3(G)$. Since $|P|=9=3^2$ (the square of a prime), $P$ must be abelian. Hints: (1) Study the N/C Theorem. Observe that $P \subseteq C_G(P) \subseteq G=N_G(P)$. Hence $G/C_G(P)$ embeds homomorphically into $Aut(P)$. (2) Note that $Aut(P) \cong C_6$ or $\cong GL(2,3)$. Hence $|Aut(P)|$ divides $6$ or $48$, so $|G/C_G(P)|$ divides these numbers, ...

2

OK, without automorphisms: put $\{P\}=Syl_3(G)$, $Q \in Syl_5(G)$ and $R \in Syl_7(G)$. Observe that $G/P \cong C_{35}$ (I leave it to you to prove this with again Sylow Theory). Hence $QP \unlhd G$, since $QP/P \in Syl_5(G/P)$. Further it is easy to see that $Q$ is the only Sylow $5$-subgroup of $QP$. Hence $Q \text{ char } QP \unlhd G$, so $Q$ is normal in ...

2

An idea: let $\;X:=\{h_i\;:\;\;i\in I\}\;$ be a generator set of $\;H\;$, and let $\;\{g_1+H,..,g_k+H\}\;,\;\;g_i\in G\;$ be a free generator set of $\; G/H\;$ == Show that $\;x\in\langle g_1,...,g_k\}\cap\langle X\rangle\implies x=0\;$ == Show that $\;\langle\{ g_1,...,g_k\}\cup\{ X\}\rangle=G\;$

2

I will assume you know how to check that $G'$ is normal subgroup, and thus that quotient $G/G'$ is well defined. I will also assume that you know how to check that canonical map $f= (x\mapsto xG')\colon G\to G/G'$ is indeed homomorphism, and not only that, it is epimorphism. I will also assume that you know Fundamental theorem on homomorphisms (and if you ...

2

$K_4=\mathbf{Z}/2\mathbf{Z}\times\mathbf{Z}/2\mathbf{Z}$ (product of rings has naturally a ring structure, defined componentwise).

2

The question is unclear, since as Derek Holt mentioned, "solvable word problem" has not been defined. If you have a presentation with a sequence of generators, you can ask two questions: 1) whether this group endowed with this generating family has a solvable word problem 2) whether this group is isomorphic to a computable group (i.e. is finite or ...

2

If $G$ is a group of prime order, then every nontrivial element generates $G$.

2

Hint: (1) $Q_8, S_3.$ (2) $\mathbb Z.$

2

Maybe good to know if you are interested: finite groups for which the property does hold (that is, for every divisor $d$ of the order $|G|$, $G$ has a subgroup of order $d$) have been studied extensively. They are called CLT (Converse Lagrange Theorem) groups. It is known that a CLT group must be solvable and that every supersolvable group is a CLT group: ...

2

I don't know how 'real' this is, but here's one way to visualise $\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$. There are two copies of $\mathbb{Z}$, call them $\mathbb{Z}\times\{0\}$ and $\mathbb{Z}\times\{1\}$. $$\mathbb{Z}\times\{0\}\qquad :\qquad \dots\qquad -3\qquad -2\qquad -1\qquad 0\qquad 1\qquad 2\qquad 3\qquad \dots$$ $$\mathbb{Z}\times\{1\}\qquad ... 2 Here's a simple geometric interpretation. Consider the figure below, with marked fenceposts (or copies of the letter C) at every integer point on a horizontal line. Its group of symmetries is \Bbb Z\times \Bbb Z/2\Bbb Z, generated by the horizontal translation and reflection across the horizontal. 1 I'm assuming that you are asking about group automorphisms because of the notations you used [(\mathbb Z/n\mathbb Z, +), (\mathbb{Q}, +), (\mathbb R, +)]. (1). (\mathbb Z/n\mathbb Z, +): Let G be a cyclic group of order n and let x \in G be a geneator. Let \phi : G \to G be a group homomorphism. First note that, \phi is uniquely determined by ... 1 Unfortunately, your attempt at a proof isn't valid. You define n to be the order of a, then prove that a^n=1 (which doesn't need proof, actually, because that's what n does), and then derive a contradiction because you had claimed n<m. Indeed, I don't think you ever used the hypothesis that m\nmid k, other than to (correctly) derive a^k\ne1 ... 1 Subgroup lattices of finite abelian groups have been studied from various points of view, see the following reference: Vogt, Frank : Subgroup lattices of finite Abelian groups: Structure and cardinality. In: Lattice theory and its applications. Hrsg.: K.A. Baker, R. Wille. S. 241-259. Heldermann , Berlin . [Buchkapitel], (1995) 1 Suppose that \mathsf{hom}(A, -) has a left adjoint T_A. In other words you have a natural (in B, C) isomorphism$$\mathsf{Ab}(A, \mathsf{hom}(B,C)) \cong \mathsf{Ab}(T_A(B), C).$$Since \mathsf{hom}(\mathbb{Z}, C) \cong C, it follows that T_A(\mathbb{Z}) \cong A. As all left adjoint, T_A preserves colimits, so for any family \{\alpha\}, ... 1 Use free groups are projective. Theres a short exact sequence 0\to H\to G \to \mathbb Z^k\to 0 that splits. 1 Just a remark - this does hold in much greater generality. According to a theorem of Nikolay Nikolov and Dan Segal, in any topologically finitely-generated profinite group (that is, a profinite group that has a dense finitely-generated subgroup) the subgroups of finite index are open. In particular, this holds for the p-adic integers. 1 No need to refer to \mathbb Q: we have the following short exact sequence of free \mathbb Z-modules (or abelian groups, if you like it)$$0\to\ker\phi\to\mathbb Z^n\to\operatorname{im}\phi\to0$$which is split (helpful, but not necessary to conclude). 1 Proof: since no one has yet answered, I shall make an attempt and would appreciate your correction: Let us denote the basis \{x_1+x_2+4x_3=a,\ 3 \cdot (x_2+2x_3)=b\} then 3a-b=3x_2+6x_6. Then obviously, since 3x_2+6x_6\in B then the order of (x_1+2x_3)+B is 3, as (x_1+2x_3)+B\ne B, 2(x_1+2x_3)+B\ne B but 3(x_1+2x_3)+B=B= identity_{A/B}. 1 Obviously$$ ( H \cap \mathbb{Z}_n) \oplus ( H \cap \mathbb{Z}_m) \leq H$$Now suppose r, s \in \mathbb{Z} such that$$sn + rm = 1$$Suppose h \in H,$$h = a + b$$with$$a \in \mathbb{Z}_n\ \ b \in \mathbb{Z}_m $$Then$$ rmh =rm(a+b)=rma= (rm +sn)a = a This means that $a \in H \cap \mathbb{Z}_n$ because obviously $rmh \in H$. Analogously \$b \in H ...

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