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7

I claim that the colimit of the sequence of inclusions $\frac{1}{1!}\mathbb{Z}\,/\,\mathbb{Z} \hookrightarrow \frac{1}{2!}\mathbb{Z}\,/\,\mathbb{Z} \hookrightarrow \frac{1}{3!}\mathbb{Z}\,/\,\mathbb{Z} \hookrightarrow \dotsc$ exists in $\mathsf{FinAb}$. Namely, it is zero, in contrast to the colimit in $\mathsf{Ab}$, which is $\mathbb{Q}/\mathbb{Z}$. This is ...


4

Use the fundamental theorem of finitely generated abelian groups, which states that every such group is a product of cyclic groups. If you cannot use this, note that every element has either order $p$ or order $p^2$. If an element has order $p^2$, then we have a cyclic group of order $p^2$. Otherwise every element has order $p$. Choose two elements neither ...


3

Lemma :$T/Z(T)$ is cyclic implies $T$ is abelian. Proof:Proving that if $G/Z(G)$ is cyclic, then $G$ is abelian. Can you conclude now?


3

Your first paragraph is in some sense backwards: you should hurry to incorporate the Eckmann-Hilton argument into your intuition, and then you can use it to predict lots of other things, including but not limited to the fact that topological groups have abelian fundamental groups! For example, you can also use it to predict that the higher homotopy groups ...


3

It's not true for $M=C_2\times C_4$. Let $x$ and $y$ be generators of the two factors. Then every endomorphism $\varphi$ is of the form $$\begin{align} \varphi(x)&=ax+2by\\ \varphi(y)&=cx+dy \end{align}$$ for integers $a,b,c$ determined$\pmod{2}$ and $d$ determined$\pmod{4}$. But if $\varphi$ is an automorphism then $\varphi(y)$ has order $4$, and ...


2

Hint. First, you missed $\def\Z{\mathbb Z}\Z_9^2$. For elements of order 27, recall that in a direct product $G \times H$, we have for $g \in G$, $h \in H$ that $$ \def\ord{\mathop{\rm ord}}\ord (g,h) = \def\lcm{\mathord{\rm lcm}}\lcm(\ord g, \ord h) $$


2

It appears that this is correct: the only countable LCA groups are discrete. Using the first structure theorem for LCA groups, we can reduce this to the case where the group $G$ is compact. According to this, there is a theorem: "Every countable compact Hausdorff space is homeomorphic to some well-ordered set with the order topology." However, the only ...


2

There is a free abelian group of rank $n$ and an epimorphism $\pi: F \to G$. Given a subgroup $H \leq G$, its preimage under $\pi$ is a subgroup of $F$, hence is free abelian of rank $\leq n$ and so is generated by at most $n$ elements; furthermore, a basis for $\pi^{-1}(H)$ may be chosen so that the basis elements of $\ker\pi$ are multiples of the basis ...


2

The finite Abelian group $G$ is a direct product of at most $t$ cyclic groups if and only if each of its Sylow $p$-subgroups is a direct product of at most $t$ cyclic groups. Hence it is enough to consider the case that $G$ is a $p$-group, so suppose it is. Now if $G$ is the direct product of exactly $t$ cyclic subgroups, then $G_{p} = \{g \in G: g^{p} = ...


2

The issue with your proof is that canceling $a$ from the left and $b$ from the right of $$ababab=aaabbb$$ only gives you $baba=aabb$. Notice that this is different from what you got which is $abab=aabb$. Perhaps the issue for you was cancelling from the middle as Steven Stadnicki has suggested. It looks like this question has been created to answer that ...


2

Suppose $(ab)^3 = a^3b^3$, that is: $ababab = aaabbb$ Multiplying on the left by $a^{-1}$ and on the right by $b^{-1}$, we have: $baba = a^{-1}abababb^{-1} = a^{-1}aaabbbb^{-1} = aabb$ That is, $(ba)^2 = a^2b^2$. Since we know $(ab)^5 = a^5b^5$ as well: $(ab)^5 = ababababab = a(babababa)b = a(ba)^4b$, so $a(ba)^4b = a^5b^5$, and multiplying on the ...


2

If $H$ is any group then you can think of a group $G$ as a subgroup of $H \times G$ by the mapping $x\mapsto (0,x)$.


2

For any $x \in H_1, y\in H_2$, we have $$ xyx^{-1}y^{-1} \in H_1\cap H_2 $$ since both are normal. However, $|H_1\cap H_2|$ divides both $p_1$ and $p_2$, and so must be 1. Hence, $xy=yx$. Thus the map $$ H_1\times H_2\times H_3 \to G \text{ given by } (x_1,x_2,x_3) \mapsto x_1x_2x_3 $$ is a homomorphism. This gives the isomorphism you need.


2

The cyclic group generated by an element is always Abelian (since anything commutes with itself, so powers of an element commute with powers of the same element). This is the cyclic group generated by $\sigma$ so it is abelian.


1

Hint : Assume $a,b$ are non-commuting elements of $S_4$. What can you say about the elements $\{0\} \times a$ and $\{0 \} \times b$ ? Remember how is defined the group law on the product. As @MPW said, the same argument works for a product $\prod_{a \in A}G_a$ of groups, when only one group is non-abelian.


1

When we have $xyz=x'yz'$ and cancel out the middle factor, we can only do that if we know that the commutes with $x$ and $x'$, or with $z$ and $z'$. From being so accustomed to working with commutativity, we lose sight of this fact. In reality, we can only cancel from the left or right, and the only way we can do more is if we can move a factor to the left ...


1

There is an equivalent of Euler's theorem for polynomials: Theorem: (Euler's Theorem for Polynomials). Let $m \in K[x]$, where $K$ is a finite field, and let $$\varphi(m) = \#\{f \in K[x] : 0 <= deg(f)< deg(m) \text{ and } gcd(f, m) = 1\}$$ Where here $\#(A)$ denotes the cardinal of the set $A$ up to a multiplication by a constant ($f$ and ...


1

The proof that the multiplicative group of a finite field is cyclic uses the property of a cyclic group $C$ that for every natural number $n > 0$, there are at most $n$ elements in $C$ of order $n$. This is an elementary (first-order) statement, for each $n$, which also holds for the infinite cyclic group (vacuously). For the finite cyclic groups, it is ...



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