# Tag Info

## Hot answers tagged abelian-groups

7

The statement is false, here's a counterexample: Let $G = (\Bbb Z,+)$, which is an abelian group. Define $T(x) := -x$, which is indeed a permutation on $\Bbb Z$. For any $x,y \in \Bbb Z$ with $x \neq y$, we have $$x-T(x) = 2x \neq 2y = y - T(y).$$ But $p(x) = x - T(x) = 2x$ is not a permutation on $\Bbb Z$ since $1$ is not in its image.

7

Hint: $\mathbb Q$ has the following property: for any two nonzero elements $a,b$ we can there exist natural numbers $n,m$ such that either $$\underbrace{a+\dots+a}_n=\underbrace{b+\dots+b}_m$$ or $$\underbrace{a+\dots+a}_n=-\left(\underbrace{b+\dots+b}_m\right)$$

5

Let $X_k$ be the set of elements of $G$ of order $k$. We know that $X_3,X_5$ are non-empty. Assume $X_{15}=\emptyset$. As distinct subgroups of prime order have trivial intersection, we conclude that $|X_5|$ is a multiple of $4$ and $|X_3|$ is a multiple of $2$. Thus $|X_3\cup X_5|=14$ is only possible if $(|X_3|,|X_5|)$ is one of $(2,12)$, $(6,8)$, $(10,4)$....

4

Alternative proof : $a^2=1$ so every element is its one inverse. So, $(ab)^{-1}=ab$, but $(ab)^{-1}=b^{-1}a^{-1}=ba$ by using twice again the remark.

4

Your proof is correct. You can improve it by making the derivations clearer (as you have done in the comments).

4

Let $G$ be a group of order $15$. There exist an element $a$ of order $5$ and an element $b$ of order $3$. Now we consider $b^{-1}ab$, and we have $(b^{-1}ab)^5 = 1$. So $o(b^{-1}ab)\mid 5$, which implies $o(b^{-1}ab) = 1$ or $5$. It cannot be $1$, otherwise $ab = b$ and so $a = 1$, a contradiction. Thus $o(b^{-1}ab) = 5$. We claim that $b^{-1}ab = a^m$ for ...

3

Fact. If $G/Z(G)$ is cyclic then $G$ is abelian. For a proof, see the answers to this old question. As in the question $G$ is non-abelian, the above fact means that $G/Z(G)$ is non-cyclic. Suppose that $Z(G)$ is non-trivial. Then $G/Z(G)$ has order either $2$ or $5$ (why?). Hence, $G/Z(G)$ is cyclic (why?), a contradiction. Hence, $Z(G)$ is trivial as ...

3

Any non-abelian group of order $2p$ is isomorphic to the dihedral group $D_p$. For $p=5$ we obtain $D_5$, which has trivial center. References: Looking for a simple proof that groups of order $2p$ are up to isomorphism $\mathbb{Z}_{2p}$ and $D_{p}$ . Center of dihedral group

3

Sure. Instead of Sylow's theorems we can get normality of a subgroup by the following: if $p$ is the smallest prime dividing $|G|$ then any subgroup of index $p$ is normal in $G$ (you can consider this an exercise if you want, but a proof can be found here). For us, $|G|=15$ so $p=3$. Now, there exists an element $g\in G$ of order $5$ (why?). This generates ...

3

Since $\{\emptyset\}$ isn't even a subgroup of $\left\langle\hat{\frac16}\right\rangle$, it cannot possibly be a subgroup. Besides,$$\left\langle\hat{\frac16}\right\rangle=\left\{\hat{\frac06},\hat{\frac16},\hat{\frac26},\hat{\frac36},\hat{\frac46},\hat{\frac56}\right\}$$Note that $\left\langle\hat{\frac16}\right\rangle$ is a cyclic group of order $6$. ...

3

If additive groups $A$ and $B$ are $\mathbb Q$-vector spaces, then any group homomorphism $f : A \rightarrow B$ is actually $\mathbb Q$-linear, since for $a \in A$ and $n \in \mathbb N$, we have $n\cdot f({a\over n}) = f(a)$ by properties of group homomorphisms, so $f({a\over n}) = {f(a) \over n}$ because $nx=y$ has a unique solution in $\mathbb Q$-vector ...

3

The exercise is wrong. The exercise seems to imply that $$\Bbb{Z}\times(\Bbb{Z}/10\Bbb{Z}) \cong(\Bbb{Z}\times\Bbb{Z})/\langle(0,1)\rangle,$$ which is false. Instead this should be $$\Bbb{Z}\times(\Bbb{Z}/10\Bbb{Z}) \cong(\Bbb{Z}\times\Bbb{Z})/\langle(0,10)\rangle,$$ which you should prove for yourself, if this is not yet clear. Then accordingly, the first ...

2

Look at $(\mathbb{Z} /2\mathbb{Z})^n$ as a vector space over $\mathbb{Z} /2\mathbb{Z}$. Then every subgroup of index $2$ is a subspace of dimension $n-1$ and so is isomorphic to $(\mathbb{Z} /2\mathbb{Z})^{n-1}$.

2

If you take a step back, then your group can be characterized as being a finite group $G$ with the property that $g^2=1$ for all $g \in G$. This property is inherited by subgroups and quotients.

2

This is actually a general fact about normal subgroups. Let $G$ be a group with normal subgroup $H$, and let $S\subseteq G$ be a maximal set of distinct coset representatives for $H$. Then $G$ is the internal direct product of $H$ and $\langle S\rangle$. Since $G$ is the union of cosets of $H$, this means that any $g\in G$ lies in some $sH$ for $s\in S$. ...

2

This is exactly the reason why I prefer to denote the identity of a general group by $e$. You can't confuse it with anything. In most books it is denoted by $e$ by the way. Anyway, you want to show that $G=K+\langle g\rangle$. First of all it is clear that $K+\langle g\rangle\subseteq G$, because a sum of an element in $K$ and an element in $\langle g\... 2 The only idea that I have is the following: If G is a group that verifies the nullity+rank theorem, i.e. for all morphism$\psi: G\to G$then$G$can be written as$G\cong ker(\psi)\oplus im(\psi)$, then each injective morphism is surjective. In general your question is really complicated. For example in a Banach space$X$if you have a morphism$T$such ... 1 A nice trick is Burnside's$p^aq^b$theorem: it says that if$p,q$are distinct primes and$G$is a group of order$p^aq^b$,$a,b\geq 1$, then$G$must be solvable. But a nonabelian simple group cannot be solvable: its commutator subgroup$G'$is normal, so$G=G'$, which means the lower central series cannot terminate. In this way you can rule out a lot of ... 1 Since$G$is a finite abelian group, we can write $$G=\langle g_1\rangle\times\cdots\times\langle g_k\rangle$$ as a direct product of cyclic groups, of orders$n_1,...,n_k$, respectively. Letting$n=\text{lcm}(n_1,...,n_k)$, it follows that$g^n=1$for all$g\in G$. Thus, for$g\in G$, if$m=\text{ord}(g)$, then$m|n$. Now let$h=g_1\cdots g_k$, and let$...

1

I believe the result is actually that $\mathbb{Z}\otimes Ag\cong Ag$. To prove this, we just need to define two homomorphisms $f\colon\mathbb{Z}\otimes Ag\to Ag$ and $h\colon Ag\to\mathbb{Z}\otimes Ag$ which are inverses of one another. Well, then let's consider: $$f(n,g)=ng\quad\text{and}\quad h(g)=(1,g)$$ First notice that $f$ is bilinear so it's well-...

1

Here is the key in induction: If $G$ is a finite abelian group of order $n=ab$, with $a,b >1$ and $\gcd(a,b)=1$, then $G \cong A \times B$, where $A= \{ x \in G : ax = 0 \}$ and $B= \{ x \in G : bx = 0 \}$. The map $A \times B \to G$ is given by $(\alpha,\beta) \mapsto \alpha+\beta$. Its inverse is given by $x \mapsto (bvx, aux)$ where $au+bv=1$. The ...

1

We'll induct on the number of distinct prime factors in the group's order. As you've said, the base case is pretty easy. Now, suppose the statement has been proven for all abelian groups, $H$, such that $|H|$ has $n$ distinct prime factors. Let $G$, then, be an abelian group with $n+1$ distinct prime factors. This means that we can factor $|G|$ as $p^kj$ ...

1

HINT: The nonzero ideals of $\Bbb{Z}_p$ are of the form $p^n\Bbb{Z}_p$, so if the induced map is not injective then $\hat{\phi}(p^n)=0$ for some $n\in\Bbb{N}$.

1

Consider the module homomorphism $$f:S\oplus T\to M$$ $$f(x,y)=x+y$$ Denote by $\Delta=\{(x,-x)\ |\ x\in S\cap T\}$ and note that $f(x,y)=0$ if and only if $x=-y$ and so $f(x,y)=0$ if and only if $(x,y)\in\Delta$. It follows that $$Im(f)\simeq (S\oplus T)/\Delta$$ Now since every finite $A$-module has to have cardinality of the form $p^x$ then $|S\cap ... 1 Side note: It would probably be a good idea to say what$n,b,c$are. I'm assuming$n\in \Bbb N$,$b\in B$and$c\in C$, with$b+c$being the element$(b,c)\in B\oplus C$. Or something similar. Maybe$b\in B\times \{0\}$and similarly for$c$makes more sense, but the result is the same. If$n\mid (b+c)$, then there is an$(x,y)\in B\oplus C$such that$n(x,...

1

In this answer, I explain how $(\beta \cdot (a \otimes b))(x \otimes y) = \beta( (ax) \otimes (by))$ makes sense. Those are equalities (1) and (2) below. First I want to isolate two claims we will use. These are things we should sort out first if we want to answer the question. For the following two claims, I want to emphasize that I am not using the ...

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