# Tag Info

8

The obvious homomorphism $\phi: G\to G/H\times G/K$ is injective, since the kernel is the set of elements in $H\cap K$. Since both quotients are abelian, the product is abelian, and $G$ is isomorphic to a subgroup of the product, and is thus abelian itself.

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 ...

5

Hint: $G$ is abeliean, and every element $a_i$ has some inverse $a_j$.

5

It's not even true for abelian groups in general. Take $H=C_2\times C_2$ as a subgroup of $G=C_4\times C_2$. Then $|Aut(G)|=8$ while $|Aut(H)|=6$.

4

The Klein group is abelian and order four, but no element has order four. More generally, a group that is the direct product of two cyclic groups that have orders that are not coprime will not have an element of the order of the group.

3

Since $G$ is abelian, we can rearrange the order $a_n$ such that $x=a_n^{-1}\cdots a_2^{-1}a_1^{-1}$. Hence, $x\cdot x=(a_1a_2\cdots a_n)\cdot(a_n^{-1}\cdots a_2^{-1}a_1^{-1})=e$.

3

Let $a, b \in G$. Consider $[a, b]:=aba^{-1}b^{-1}$ which is frequently called the commutator. (Up to choice of convention.) Then the hypotheses tell us that $[a, b] \in H$ and $[a, b] \in K$ because $ab=ba$ mod $H$ or $K$.

3

The Smith Normal Form of your matrix is $\left(\begin{smallmatrix}1&0\\0&9\end{smallmatrix}\right)$ hence $A\cong \mathbb{Z}/9\mathbb{Z}$. More generally, the SNF will be a block diagonal matrix $\left(\begin{smallmatrix}I&0&0\\0&D&0\\0&0&0\end{smallmatrix}\right)$. Ignore the first block of columns and rows, leaving ...

2

Let us write $G'$ for your $C$, which is more common in group theory. First note that $G'=\{e\}$ if and only if $G$ is abelian. This can be easily seen: if $a, b \in G$, then $ab=ba$ if and only if $[a,b]=e$. Now what you are asking is if $G'$ itself is abelian. What then? Well, according to the previous remarks, then $(G')'$, written as $G''$ must be ...

2

No. A finite abelian group $G$ is never a free $\mathbb{Z}$-module, because $\mid G\mid n=0$ for all $n$. Yes, $\mathbb{Z}$ itself, for example, is a free $\mathbb{Z}$-module, with basis $\{ 1\}$.

2

The stupid answer is that $\operatorname{Hom}_{\text{cts}}(\mathbb Z, S^1) = \operatorname{Hom}_{\mathbb Z}(\mathbb Z,S^1) = S^1$, so we are done by duality. Here, $S^1$ denotes the unit circle $S^1 \cong \frac{\mathbb R}{2\pi\mathbb Z} \cong \mathbb R/\mathbb Z$, a.k.a. $\mathbb T$. The real answer is that all characters of $S^1$ are of the form $z ... 2 I would reason like this. A set of representatives for your quotient is $$\{(0,n,m)+\langle(1,1,1)\rangle\ |\ n,m\in\mathbb{Z}\}$$ since$(a,b,c)+\langle(1,1,1)\rangle=(a-a,b-a,c-a)+\langle(1,1,1)\rangle$so $$\frac{\mathbb{Z}\times\mathbb{Z}\times\mathbb{Z}}{\langle(1,1,1)\rangle}\cong\mathbb{Z}\times\mathbb{Z}$$ Can you show$\mathbb{Z}\times\mathbb{Z}$... 2 This is true without any f.g. assumption and is equivalent to the statement that the circle group$U(1)$is injective, which follows from Baer's criterion since it is divisible. If your groups are f.g., then you won't need Zorn's Lemma in Baer's criterion (it can be replaced by a simple induction). 2 Yes. Assuming the axiom of choice the answer is positive. You can find the proof in W.R. Scott's paper: Scott, W. R. "Groups and cardinal numbers." Amer. J. Math. 74, (1952). 187-197. The axiom of choice is used there for all manner of cardinal arithmetics. Without the axiom of choice it is no longer necessary that the proof can go through. Because ... 1 what would be the order of any element in$\Bbb Z_8^*=\{1,3,5,7\}$, so there is no element of order$4$1 First, note that ii and iii are equivalent since$Im(\phi)\cong G/Ker(\phi)$. To prove this, just take the subnormal series for$G$and project it onto$G/N$, which should preserve normality and abelianness of consecutive quotients (you can check that the commutator of$G_{i+1}/N$is contained in$G_i/N$). For iv, since subgroups of$G/N$correspond to ... 1 Take two character$\chi_1$and$\chi_2$are in$\widehat{G}$. Take$H:=H_w$where$w$is a fixed$f(p)$root of unity. First assume that$\chi_1$and$\chi_2$are both in$H$then the character$\chi_1\chi_2^{-1}$verifies : $$\chi_1\chi_2^{-1}(p)=\chi_1(p)\chi_2(p)^{-1}=ww^{-1}=1$$ So$\chi_1\chi_2^{-1}\in U$. By definition of coset of$\widehat{G}$mod ... 1 See: http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/cauchyapp.pdf To be exact, see theorem 1.1, it relies on Cuachy's Theorem. As already pointed out, the "abelian" part is not necessary. 1 By the fundamental theorem for finitely generated modules over a p.i.d., a quotient group of$\mathbf Z^2$will be cyclic if and only it has only$1$invariant factor$\neq 1$. 1 Sorry, I completely misread the question. For some reason I thought you wanted$B \le A$. Let's try again. I think the answer is yes. Let$H$be the subgroup of$G$generated by all images of$A$under automorphisms of$G$. Then$G:H$is finite and$H\ {\rm char}\ G$. But, since$H:A$is finite,$H$must be generated by finitely many of the images ... 1 The order of any element in direct product of groups is determined as Let$G$=$G_1 \times G_2 \times \cdots \times G_n$. Then order of$(a_1, a_2 , \cdots, a_n)\in G$is LCM of orders of$a_1, a_2, \cdots, a_n$,$i.e.$, $$\vert (a_1, a_2 , \cdots a_n) \vert = \texttt{lcm }(\vert a_1\vert, \vert a_2\vert, \cdots , \vert a_n\vert ).$$ Every ... 1 No, this would imply that all (infinite) countable groups are isomorphic (as the underlying sets of any two such groups are in bijection), but that is not the case: For example both$(\mathbb{Z}, +)$and$(\mathbb{Z} \times \mathbb{Z}, +)$are countable, but they are not isomorphic, because the former is generated by a single element but the latter is not. 1 The following map suggests itself as candidate for an isomorphism$\phi\colon K\to H\times C_2$: $$\phi(x)=\begin{cases}(x,0)&\text{if x\in H}\\(gx,1)&\text{if x\in gH}\end{cases}$$ Check it. For part 4, note that$G$contains some subgroups of the form$C_2^k$(for example the trivial subgroup, where$k=0$). Among those subgroups, let ... 1 Maschke's theorem says that$\Bbb C[\Bbb Z/(n)]$is a semisimple ring. Since the cyclic group is abelian, we are talking about a commutative$\Bbb C$-algebra, and as a corollary of the Artin-Wedderburn theorem, a commutative semisimple$\Bbb C$-algebra is a product of finite field extensions of$\Bbb C$. Since$\Bbb C\$ is algebraically closed, there are ...

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