# Tag Info

7

No this is not true in general. Let $G = \langle a \rangle \times \langle b \rangle$ with $|a|=2$, $|b|=4$. Then $\{ ab,b \}$ is a minimal generating set (in any sense), but they both have order $4$, whereas the group has order $8$. Of course by the Fundamental Theorem, there always exsists a minimal generating set with the property you describe, such as ...

5

It is not hard to show that $G(m)$ and $G(k)$ are subgroups and intersect each other trivially. We need to show then that $G(m)G(k)=G$. Suppose $g\in G$ is an element; then since $m$ and $k$ are relatively prime we have that there are integers $a,b$ such that $am+bk=1$. Then $g=g^{am}g^{bk}$, and $g^{am}\in G(k)$ and $g^{bk}\in G(m)$, so $G(m)G(k)=G$.

4

HINT: you need to show that $g^{nd}h^{nd}=h^{nd}g^{nd}$ and $g^{cm}h^{cm}=h^{cm}g^{cm}$.

4

Let $g_1,g_2\dots g_n$ be the generators for the group, then every element of the group can be seen as $g_1^{a_1}g_2^{a_2}\dots g_n^{a_n}$ where $0\leq a_i< |g_i|$. This tells us $|G|\leq\prod\limits_{i=1}^n|g_i|$. The inequality can be sharp however, consider the group $\mathbb Z_4+\mathbb Z_6$ of order $24$. The set $\{(1,2),(1,1)\}$ is a minimal and ...

3

This doesn't hold in general (for example, we have $\mathbb{Z}^{\mathbb{N}} \cong \mathbb{Z}^{\mathbb{N}} \oplus \mathbb{Z}^{\mathbb{N}} \cong \mathbb{Z} \oplus \mathbb{Z}^{\mathbb{N}}$), but under some conditions it does hold: Krull-Schmidt theorem.

3

The number of non-isomorphic Abelian groups depends on the partition number of the exponents in the prime factorization. Let $P(k)$ be the number of ways to partition $k$ (for example, $P(4) = 5$, since we have $4, 3+1, 2+2, 2+1+1$, and $1+1+1+1$). Then for $n=p_1^{e_1} \ldots p_r^{e_r}$, there are $$\prod_i P(e_i)$$ non-isomorphic Abelian groups of order ...

3

You can simplify things a lot by noting that $|a|$ divides $K$ iff $a^K=1$. Using this, verifying that $H$ is a subgroup is easy. If you can use homomorphisms, then it is easier because $H=\ker\phi$, where $\phi(x)=x^K$. Since $G$ is Abelian, $\phi$ is a homomorphism.

2

To amplify on on Yotas's comment: look at $S_3$, the symmetries of an equilateral triangle. The subgroup generated by "flip about a median" has order 2; rotation has order 3. If these are $H$ and $K$, then $G = HK = KH$, but the elements of $K$ and $H$ do not commute, so $G$ is not abelian. Hence your second highlighted paragraph is mistaken.

2

Hint One can show directly that the operation $\ast$ is conjugate to the usual addition operation on $\Bbb R$ via the hyperbolic tangent function, which satisfies the suggestive identity $$\tanh (s + t) = \frac{\tanh s + \tanh t}{\tanh s \tanh t + 1}.$$ Alternate hint One can rearrange the desired inequality $x + y < xy + 1$ (note we've dropped the ...

2

The order of this group is equal to the number of spanning forests of the graph. Compute the 2-sylow subgroup of this group and look at the number of cyclic factors in a direct sum decomposition; this number is the dimension of the so-called "bicycle space" of the graph (bicycles are subgraphs that are both edge-cuts and have even degree at each vertex; ...

2

It does not really nswer the question you originnaly posted, but here's a hint for the second question if you're looking for one : (I hope this works) To tackle the second question you posted, I can give you a hint : Consider $G=\mathbb{Z}^{2}$. Now we are looking for subgroups H of index $p$, hence $G/H$ is a group of order $p$ hence cyclic. For that, we ...

2

First, obviously, this operation is commutative on $\Bbb R\times\Bbb R$. Second, also obviously, it is associative on $\Bbb R\times\Bbb R$. You need to show that there exists a neutral element $e$; in other words, $$\forall x\in \Bbb R\quad \arctan (\tan x+ \tan e) = x,$$ which implies - after taking $\tan$ of both sides - that $\tan e =0$. Can you ...

2

Let $e$ be the identity element, then \begin{align*} x * e & = x\\ \arctan(\tan(x) + \tan(e))& =x\\ \tan(x) + \tan(e) & = \tan (x)\\ \tan(e)&=0. \end{align*} This suggests that $e=0$ can act as identity.

2

Hint: you just need some $e$ with $\tan(e)=0$, then you have $x*e = \arctan ( \tan x + 0 ) = x$.

2

Because $\mathbb{Z}[i]$ is a PID, we have Bezout's identity (Wikipedia link), so we can see that $a+bi\in\mathbb{Z}[i]$ becomes a unit when considered in $\mathbb{Z}[i]/(8)$ iff $\gcd(a+bi,8)=1$. The factorization of $8$ in $\mathbb{Z}[i]$ is $$8=2^3=i(1+i)^6$$ so $a+bi$ becomes a unit modulo $8$ iff $\gcd(a+bi,1+i)=1$. Now plot elements of $\mathbb{Z}[i]$ ...

2

Note that every abelian group can be seen as a $\Bbb{Z}$-module and that a $\Bbb{Z}$-linear map between abelian groups is just a group homomorphism. Then this answer shows you how to prove that $A \otimes (B \otimes C) \simeq (A \otimes B) \otimes C$ as modules, hence as groups, where the isomorphism is the obvious one: $$\varphi \colon \;\, a \otimes (b ... 1 It’s only true when x and y are restricted to the interval \langle-\pi/2,\pi/2\rangle. For when they are so restricted, \tan and \arctan become two-sided inverses of each other. Once you do that, you’ve just transformed ordinary addition on the open interval \langle-\infty,\infty\rangle into the new addition on the bounded interval. 1 Set H:=\langle (q(A))^{K/N}\rangle  and take q(k)q(a)q(k)^{-1} be any element of  (q(A))^{K/N} then clearly we have that q(k)q(a)q(k)^{-1}=q(kak^{-1})\in q(\langle A^K\rangle). Hence a generating set of H is in q(\langle A^K\rangle), from this it follows that H\subseteq q(\langle A^K\rangle). If \langle A^K\rangle is abelian, so is ... 1 Yes. This follows because the image of \langle A^K\rangle is a normal abelian subgroup of K/N which contains the image of A, and hence also contains the conjugate closure of the image of A (recall that the conjugate closure of a subset is the smallest normal subgroup containing that subset). 1 \mathbb C^{\times} is an Abelian group. The map z \mapsto |z| is a group homomorphism \mathbb C^{\times} \to \mathbb R^{\times}. The circle is the kernel of this map and so is a subgroup of \mathbb C^{\times}. The crucial point is that |zw|=|z|\ |w| and this can be proved as follows:$$|zw|^2=(zw)(\overline{zw})=z w \bar z \bar w=z \bar z w \bar ...

1

Hint: if $z\in G$, there exists $\varphi\in(-\pi,\pi]$ with $z=e^{i\varphi}$. (Of course one can choose $\varphi \in [0,2\pi)$ or $\varphi \in I$ for any half closed intervall $I$ of length $2\pi$).

1

Write $mk+nl=1$ for some $k,l\in\mathbb Z$. We have $$ab=a^{mk+nl}b^{mk+nl}=a^{mk}(a^{nl}b^{mk})b^{nl}\stackrel{(*)}=a^{mk}(b^{mk}a^{nl})b^{nl}=(a^{mk}b^{mk})(a^{nl}b^{nl})=(b^{mk}a^{mk})(b^{nl}a^{nl})=b^{mk}(a^{mk}b^{nl})a^{nl}\stackrel{(**)}=b^{mk}(b^{nl}a^{mk})a^{nl}=(b^{mk}b^{nl})(a^{mk}a^{nl})=b^{mk+nl}a^{mk+nl}=ba,$$ where we used $(*)$ ...

1

Quick answer : no. Long one : Assume that such a subgroup exists and take two elements a and b such that ab =/= ba exist, then the result would be the same in the abelian group, which wouldn't be abelian after all.

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