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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-...

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

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

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 ...

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 $p^kj$ ...

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.

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 ...

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

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 $|S\cap ... 4 Alternative proof :$a^2=1$so every element is its one inverse. So,$(ab)^{-1}=ab$, but$(ab)^{-1}=b^{-1}a^{-1}=ba$by using twice again the remark. 4 Your proof is correct. You can improve it by making the derivations clearer (as you have done in the comments). 3 Since$\{\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 ... 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 $$\underbrace{a+\dots+a}_n=-\left(\underbrace{b+\dots+b}_m\right)$$ 0 Well, the zero divisors of${\Bbb Z}_{p^k}$are exactly the nonzero elements$a$with$\gcd(a,p^k)\ne 1$, i.e., the zero divisors are exactly the nonzero multiples of$p$. Then the$p^{k-1}$-multiple of a zero divisor$a$written as$a=pb$is$p^{k-1}\cdot a = p^{k-1}pb = p^kb=0$. 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 ... 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 ...

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

More generally, let $p$ be any prime. If one has a finite group $G$ of order $p^n$, then it has a composition series with quotients of order $p$, that is there is a sequence ${e}=G_0\triangleleft G_1\triangleleft G_2\triangleleft \cdots \triangleleft G_{n-1} \triangleleft G_n=G$ with each $|G_j|=p^j$. Each $|G_j:G_{j-1}|=p$. If $G$ is non-Abelian, then ...

5

This is, in fact, a true statement. There are two essential facts that I use here without proof. Any finite abelian group is a product of cyclic groups. Look up the structure theorem for modules over a PID for more on this. For any $n \geq 2$ there are infinitely many primes $p$ such that $p \equiv 1 \text{ mod } n$. This is a special case of Dirchlet'...

7

Yes. The Klein subgroup $\{1, (1~2)(3~4), (1~3)(2~4), (1~4)(2~3)\} \triangleleft A_4$. This subgroup is abelian, as the title of the question requests, but it's not cyclic.

1

By the Fundamental Theorem of Finitely Generated Abelian Groups, the only non-cyclic abelian group of order $63$ is (up to isomorphism) $\mathbb{Z}_{21} \times \mathbb{Z}_3$ You have successfully found that there are $8$ elements of order $3$, so now look at the subgroups generated by these particular elements. You will notice that you will find that some ...

2

As @DavidMolano commented you can view $\frac{1}{4}\mathbb{Z}$ as a subgroup of $\mathbb{Q}$ containing $\mathbb{Z}$. Two elements $\frac{a}{4},\frac{b}{4}\in\frac{1}{4}\mathbb{Z}$ with $a,b\in\mathbb{Z}$ are congruent $\mod\mathbb{Z}$ iff $4|a-b$, and thus there is an isomorphism $\frac{1}{4}\mathbb{Z}/\mathbb{Z}\to\mathbb{Z}/4\mathbb{Z}$ defined as $\frac{... 2 Suppose that$G=S_1\times\cdots\times S_k$is a direct product of nonabelian simple groups and let$N\triangleleft G$. I claim that$N$is the direct product of some of the$S_i$; that is,$N=N_1\times\cdots\times N_k$, where$N_i$is either trivial or equal to$S_i$for each$i$. Note that if$\pi_j\colon G\to S_j$is the projection on the$j$th coordinate, ... 1 The function$X\times X\to Y$prescribed by: $$(x_1,x_2)\mapsto\alpha(x_1\otimes x_2)$$ is bilinear. Conversely if$f:X\times X\to Y$is bilinear then a unique$\mathbb Z$-module homomorphism$\alpha: X \otimes_{\mathbb{Z}} X \rightarrow Y$exists such that: $$f(x_1,x_2)=\alpha(x_1\otimes x_2)$$ So there is a one-to-one correspondence between bilinear ... 5 Huh, funny, we just went over this today in my algebra class. Yes, subgroups of abelian groups are indeed abelian, and your thought process has the right idea. Showing this is pretty easy. Take an abelian group$G$with subgroup$H$. Then we know that, for all$a,b\in H$,$ab=ba$since it must also hold in$G$(as$a,b \in G \ge H$and$G$is given to be ... 2 If$G$is an abelian group and$H$is a subgroup, suppose$x, y\in H$. Then in particular$x, y\in G$, so$xy=yx$. Since$x, y$were arbitrary,$H$is abelian. 3 A slightly more general result is contained in the paper Yun Fan, A characterization of elementary abelian$p$-groups by counting subgroups. (Chinese) Math. Practice Theory 1988, no. 1, 63–65. I understand from the MathSciNet review that in this paper it is proved that among all finite groups of order$p^{n}$, where$p$is a prime, the elementary ... 4 For the second one it's really easy to see where algebraic closure intervenes : you pick an eigenvalue for one of the elements of the group : it might not exist if the field is not algebraically closed ! For the first one you have to recall where this result on the sums of squares comes from : it comes from the fact that the group algebra$k[G]$decomposes (... 1 To reiterate what was said in the comments by @ArturoMagidin, you used the fact that$G$is abelian in this step:$\therefore ab^{-1} \in S \iff {(ab^{-1})}^2 \in H$,$\therefore ab^{-1} \in S \iff a^2(b^2)^{-1} \in H. Thusly: \begin{align} (ab^{-1})^2&=(ab^{-1})(ab^{-1}) &\\ &=a(b^{-1}a)b^{-1}& \\ &=a(ab^{-1})b^{-1} & \... 0 A proof using the second and third isomorphism theorems goes like this \frac{H_{1} N}{H_{2} N} = \frac{H_{1} H_{2} N}{H_{2} N} \cong \frac{H_{1}}{H_{1} \cap H_{2} N} \cong \frac{H_{1} / H_{2}}{(H_{1} \cap H_{2} N) / H_{2}}, $$and the latter is abelian, as an homomorphic image of the abelian group H_{1}/H_{2}. (Note that H_{2} is indeed contained in ... 0 Your idea is good. After you have proved that H_2N is a normal subgroup of H_1N, you can consider the map$$ H_1\to H_1N/H_2N,\qquad x\mapsto xH_2N $$Prove this map is a surjective homomorphism. The kernel is H_1\cap H_2N. Since H_1\cap H_2N\supseteq H_2, … 1 This can be proven using only the definition of the quotient. Suppose H_1/H_2 is abelian, then for any a,b \in H_1 we have aba^{-1}b^{-1} =h \in H_2. Now let an,bm\in H_1N be given. We have$$\begin{align} anbmn^{-1}a^{-1}m^{-1}b^{-1} & = an(a^{-1}a)bmn^{-1}a^{-1}(b^{-1}b)m^{-1}b^{-1}\\ & = n'abmn^{-1}a^{-1}b^{-1}m'\\ & = n'abmn^{-1}[(ab)^... 2 LetI$be an injective abelian group and let$T(I)$be its torsion subgroup. Then$T(I)$is injective. Indeed,$T(I)$is divisible since$I$is divisible (and if$nx\in T(I)$then$x\in T(I)$). So, if$M$is torsion, then it embeds into the torsion subgroup of$F(M^\vee)^\vee$, which is still injective. Alternatively, it is pretty easy to directly prove ... 4 No, it is not true. Consider$\mathbb{Z}_2\oplus\mathbb{Z}_3$. This has a generator$(1,1)$. Note that $$0\oplus\mathbb{Z}_3<\mathbb{Z}_2\oplus\mathbb{Z}_3 ,$$ and $$(\mathbb{Z}_2\oplus 0)\oplus(0\oplus\mathbb{Z_3})=\mathbb{Z}_2\oplus\mathbb{Z}_3.$$ However,$0\oplus\mathbb{Z}_3$is generated by$(0,1).\$

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