22 votes

Name of algebraic structure (group-like)

This is a commutative quasigroup. However, personally I don't find this sort of naming to be particularly illuminating. To my mind the simplest way to think about this operation $\star$ is actually as ...
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18 votes
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(Why) Is there no analogue to the classification of finitely generated abelian groups for abelian groups?

No, it is badly false and the classification of abelian groups is much more complicated. For example $\mathbb{Q}$ is not a nontrivial direct sum in any way. Exercise 1: Prove that if an abelian group ...
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12 votes

Name of algebraic structure (group-like)

The structure you are describing is called a Steiner quasigroup. The phrase Steiner quasigroup is sometimes abbreviated 'squag', so some folks call these objects squags. In general, a (finite) Steiner ...
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8 votes
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Motivation in considering abelian groups in modules

Let $G$ be a group, $R$ a ring and assume there is an "$R$-module structure on $G$" (something satisfying the usual axioms except we don't require $G$ to be abelian). Then, let $x,y\in G$ be ...
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  • 7,930
7 votes
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Morphism of free groups that induces isomorphism on abelianizations

The alternating group $A_5$ is generated by $a = (12345)$ and $b = (12)(34)$. These give a transitive action of the free group $F_2$ on $ \{ 1, 2, 3, 4, 5 \}$. The stabilizer of $5$ is a subgroup $H$ ...
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7 votes
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Show there exists a non-abelian group of order $pq$ without semi-direct product

$\mathbb Z_q$ is a finite field, so its multiplicative group $\mathbb Z_q^{\times}$ is a cyclic group of order $q-1$. Since $p\mid q-1=|\mathbb Z_q^{\times}|$, there is a unique subgroup $P\leq \...
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7 votes
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Find an homomorphism $\phi:\mathbb{Z}^4 \to G$

The fact that $\mathbb Z^4$ is free abelian does not imply that there exists a unique homomorphism $\mathbb Z^4 \to G$. Instead, fixing the free basis of size $4$ that you indicated and that I will ...
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  • 102k
6 votes

Classification of subgroups of $(\mathbb Q,+)$ that are not finitely generated.

Let $P$ be the set of all prime numbers and let $f: P\to\mathbb{N}\cup\infty$ be an arbitrary function. Let $$ N_f=\{n\in\mathbb{N}\mid n=\prod_{p\in P} p^{k_p}, k_p\leq f(p)\}. $$ Let $\mathbb{Q}_f=\{...
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  • 5,258
6 votes

Let $G = \Bbb Z \times \Bbb Z$ with group law given by addition. Let $H$ be the subgroup generated by $(2,3)$. To which group is $G/H$ isomorphic to?

As we've noted, the kernel of your map is strictly larger than your $H$: your $H$ consists only of those elements in which the first entry is a multiple of $2$, and the second entry is the same ...
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6 votes
Accepted

For any $n$, let $\alpha(n)=$ #non-isomorphic Abelian groups of order $n$. What is the maximum value of $\alpha(n)$ for $n\le 200$?

Finite abelian groups are direct sums of their Sylow $p$-subgroups, which are direct sums of cyclic $p$-groups $C_{p^k}$, and this direct sum decomposition uniquely identifies the isomorphism class. ...
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5 votes

Show there exists a non-abelian group of order $pq$ without semi-direct product

You can use a semi-direct product without saying that you are doing so, by a matrix trick. Namely, if $\mathbb Z/p$ is to (secretly?!?) act on $\mathbb Z/q$, for $x\in \mathbb Z/p$ and $y\in \mathbb Z/...
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  • 47.5k
5 votes
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How to find subgroups of a product group containing the diagonal?

The following observation seems to me useful. Let $G$ be a group, $D$ be a diagonal in $G\times G$, $K$ be a normal subgroup of $G$. The set $$ H=\{(x,y)\mid x^{-1}y\in K\} \qquad(*) $$ is a subgroup ...
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  • 5,258
5 votes
Accepted

If two finite Abelian groups posses equal number of elements of some particular order, then justify they are isomorphic.

Trivial case: In fact any group has a unique element of order $1$, i.e., $\Bbb{Z}_4$ and $K_4$ has same number of element of order $1$. Consider $\Bbb{Z}_2\times\Bbb{Z}_6$ and $\Bbb{Z}_{12}$. The ...
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  • 9,490
5 votes
Accepted

If $G/M$ and $G/N$ both are cyclic and $\gcd(|M|,|N|)=1$, then $G$ is abelian.

Consider the homomorphism $\varphi:G\to G/M\times G/N$ defined by $$\varphi(g) =(gM, gN) $$ We have $$\begin{align}\ker \varphi&=\{g\in G:(gM, gN)=(M, N) \}\\&=\{g\in G:g\in M, g\in N\}\\&=...
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  • 9,490
5 votes

Why is $\Bbb Z/4\Bbb Z\times \Bbb Z/12\Bbb Z\times \Bbb Z/40\Bbb Z$ not isomorphic to $\Bbb Z/8\Bbb Z\times \Bbb Z/10\Bbb Z\times \Bbb Z/24\Bbb Z$?

One simple way to see this is that the second group has an element $(0,5,0)$ of order $2$ and no element $g$ such that $g+g=(0,5,0).$ There is no element like this in the first group. If $(x,y,z)\in\...
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4 votes
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If both finite groups $G_1$ and $G_2$ (of equal order) possess equal number of elements of highest possible order, will $G_1\simeq G_2$ hold?

No, e.g. $\Bbb Z_8\oplus\Bbb Z_4$ and $\Bbb Z_8\oplus\Bbb Z_2\oplus\Bbb Z_2$ have an equal number of elements of order $8$.
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  • 16k
4 votes
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Prove that $S$ is a coset of some subgroup of $G$ iff $S+S-S=S.$

Since $S$ is non-empty, pick any $t\in S$. I claim that $S-t:=\{s-t:s\in S\}$ is a group, which will show the desired result. Clearly $0\in S-t$. Thus we need only show that, for any $x,y\in S-t$, we ...
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3 votes

Is this subgroup diagram of $\mathbb{Z}_{90}$ correct?

Your idea with the parallelogram can be implemented that way. In the picture, the number means the order of the subgroup. If two subgroups are connected by an edge, the higher order subgroup lies ...
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  • 5,258
3 votes

Show there exists a non-abelian group of order $pq$ without semi-direct product

It is a theorem that $$\langle a,b\mid a^p,b^q,aba^{-1}b^{-r}\rangle $$ is a presentation of a non-abelian group of order $pq$ when $p\mid q-1, r\ne1\pmod q$ and $r^p\equiv 1\mod q$. It's easy to see ...
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  • 2,732
3 votes

What are different types of mathematical structures that can the space $\mathbb{Z}$ have?

Just to answer on the measure, there is the so-called "discrete", or "counting", measure on any set, where any subset is measurable and its measure is its (possibly infinite) ...
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  • 21.9k
3 votes
Accepted

Suppose $G$ is an Abelian group and $a,b\in G$ with $|a|=60,|b|=36$. What are possible orders of $ab$?

Theorem Suppose $G$ is an abelian group, $x,y \in G$, and the orders of $x$ and $y$ are finite. Then $|xy|$ is also finite and is a divisor $\operatorname{lcm}(|x|,|y|)$. Proof Let $m = \operatorname{...
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  • 1,491
3 votes

Suppose $G$ is an Abelian group and $a,b\in G$ with $|a|=60,|b|=36$. What are possible orders of $ab$?

Your $a$ and $b$ as groups is wrong; instead, they are elements of $G$. Here is a hint: Since $G$ is abelian, we have for all $x,y\in G$, that $$ (xy)^m=x^my^m$$ for any $m\in \Bbb Z$.
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  • 39.6k
3 votes
Accepted

How to determine the invariant factors of an abelian group of order $2^6 \cdot 3^5 \cdot 5^7$

There is a fact that serves to find them algorithmically. Fact. Let $G$ be a finitely generated group, then there exists $m\geq 0$, $p_1,\ldots,p_n\in \mathbb{Z}_{+}$, distinct primes and naturals \...
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3 votes
Accepted

Determination of order of cosets in a factor group of a finite abelian group.

Let $G_1=\Bbb{Z}_{3^{10}} $,$G_2=\Bbb{Z}_{3^{7}}$ , $H=\langle (3^2, 3^3)\rangle$ Then $|3^2|_{G_{1}}=3^8$ , $|3^3|_{G_{2}}=3^4$ Let $|a|=n$ where $a=(1,0)+H$ Then $n$ is the least positive integer ...
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  • 9,490
2 votes
Accepted

Finding order of elements order-wise in product of groups.

The direct product of two abelian groups is abelian. Thus $$G= C_{8}\times C_2\times C_1\times C_1$$ is abelian. However, since $Q_8$ is nonabelian, $Q_8\times A$ is nonabelian for any group $A$. Thus ...
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  • 39.6k
2 votes
Accepted

Composition of two elements $|a|= p^m, |b|= p^n$ in abelian group, then order of composition?

Consider the group $\Bbb{Z}_4×\Bbb{Z_8}$ Let $a=(1, 0) $ and $b=(0, 1) $ . Then $|a|=\operatorname{lcm}(|1|, |0|) =\operatorname{lcm} (4, 1) =2^2$ $|b|=\operatorname{lcm}(|0|, |1|) =\operatorname{lcm} ...
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  • 9,490
2 votes

For any $n$, let $\alpha(n)=$ #non-isomorphic Abelian groups of order $n$. What is the maximum value of $\alpha(n)$ for $n\le 200$?

In addition to the nice answer of @Qiaochu Yuan, you might want to check a paper of John Conway, Heiko Dietrich and E.A. O'Brien carrying the somewhat mysterious title Counting groups: gnus, moas and ...
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2 votes
Accepted

Two affine group schemes over $k$ with isomorphic group for each $k$-algebra $S$ but not isomorphic as affine group schemes

Note that $\mu_p(S)$ and $\alpha_p(S)$ are vector spaces over $\mathbb{F}_p$ thanks to the following actions: $$ \mathbb F_p \times \mu_p(S) \to \mu_p(S), \qquad (n,s)\mapsto s^n $$ $$ \mathbb F_p \...
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  • 3,081
2 votes

Why is addition on elliptic curves defined in this particular way?

The group law on elliptic curves comes from looking at the functions of geometric surfaces, and trying to understand the behaviour of intersections. When we look at the coordinate ring of an elliptic ...
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2 votes

If $G/M$ and $G/N$ both are cyclic and $\gcd(|M|,|N|)=1$, then $G$ is abelian.

Note that if $K \unlhd G$, then $G/K$ is abelian if and only if $G' \subseteq K$. Hence above it follows that $G' \subseteq M \cap N$, but since gcd$(|M|,|N|)=1$, it follows (Lagrange - here one uses ...
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