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Find the smallest $n$ such that $S_n$ has an element of given order.

Hint $\lvert g\rvert =\rm {lcm}(|c_1\rvert, \lvert c_2\rvert, \dots, \lvert c_k\rvert) $, where $g=c_1c_2\dots c_k$ is the cycle decomposition of $g$. Write $$m=\prod_ip_i^{a_i}$$, for the prime ...
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split maximal torus construction

The subgroup $T$ is the image of the subgroup of diagonal matrices in $GL(n,q)$. MAGMA constructs $PGL(n,q)$ as a permutation group, acting on the projective line which has $(q^n - 1)/(q-1)$ points. ...
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1 vote

Show that the group contains an element of order 4

Let $G'=\langle z\rangle$, so $z$ has order $2$. Let $g,h\in G$ such that $[g,h]=z$. As you have noted, $z$ is central. One can then check that $g^h$ (the conjugate of $g$ by $h$) commutes with $g^{-1}...
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1 vote

Show that the group contains an element of order 4

According to this question and its answer, $G$ contains a nonabelian group of order $8$. Up to isomorphism there are only two such groups and they all contain an element of order $4$.
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  • 1,233
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Show that the set $H=\{f \in S_4 \mid f(4)=4\} \subset S_4$ is isomorphic with $S_3$.

$f\in H$ implies $f(4) =4$ . Hence number of permutations / functions that fixes $4$ are exactly $3! =6$ $H$ is non abelian group of order $6$ and there exactly two groups of order $6$ upto ...
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1 vote

Let $H,K$ be subgroups of $G$. Show that if $G$ has elements $x,y$ such that $xH=yK$, then $H=K$.

As noted, your reasoning is not correct. As to a proof, while you can do one with double inclusion, you can also prove it here more directly. If $xH=yK$, then $x^{-1}xH = x^{-1}yK$, so $H=x^{-1}yK$. ...
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2 votes

Groupoid embedding into a group

I would say so. Lets say a group is a one-object categorical groupoid. Under the equivalence of definitions described in the Wikipedia article an algebraic groupoid with the property $a\ast a^{-1}=b\...
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1 vote

Is the symplectic group over the rationals $\text{Sp}(2n,\mathbb Q)$ dense on the symplectic group $\text{Sp}(2n,\mathbb R)$ over the reals?

For any field $\mathbb{F}$, the symplectic group $Sp(2n,\mathbb{F})$ is generated by symplectic transvections. These are maps of the form $f_{\alpha,u}$, where $\alpha \in \mathbb{F}$, $u \in \mathbb{...
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Let $H,K$ be subgroups of $G$. Show that if $G$ has elements $x,y$ such that $xH=yK$, then $H=K$.

Is this a correct reasoning? No, as you've written it doesn't make sense, because $y^{-1}x\in G$ doesn't mean that $h\in K$. If $xH=yK$ then $xe=x=yq$ for some $q\in K$, so $y^{-1}x=q\in K$. Once you'...
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Abstract Algebra Normal Subgroups

This just comes down to basic coset manipulations. You have $$h_1K = \{ h_1 k : k \in K\}.$$ You always have $h_1 \in h_1K$. Therefore, if $h_1K = h_2K$, this means that there exists a $k \in K$ ...
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2 votes

Isomorphism between two presentations of 6 order abelian group

One can use von Dyck's Theorem to obtain the morphisms; this will save you the work of checking the maps are homomorphisms. Let $$\begin{align} G&= \langle k\mid k^6\rangle,\\ K&= \langle s,r\...
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1 vote

Outer automorphism group being a quotient group

Here's another perspective on $Out(G)$ that might be clearer: Consider the map $\phi:G\to Aut(G)$ by $\phi(g)= \phi_g$ such that $\phi_g(h)= g^{-1}hg$ for all $h\in G$. It's a good exercise to prove ...
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1 vote
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Outer automorphism group being a quotient group

$\text{Out}(G)$ is supposed to measure the "outer" automorphisms of $G$. That is, the automorphisms that aren't inner. To do this, we start with the group of all automorphisms $\text{Aut}(G)$...
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3 votes

Isomorphism between two presentations of 6 order abelian group

I think to have found it explicitly. Let's redefine the two groups like this: $$A=\langle r\mid r^6\rangle \\ B=\langle a, b\mid a^3, b^2, aba^{-1}b\rangle $$ We can define a homomorphism $\Phi = A\...
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3 votes

Application of nonfamous finite groups in computer science

Automorphism groups of codes. For example, the sporadic Mathieu group $M_{24}$ is the automorphism group of the extended binary Golay code.
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2 votes
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Application of nonfamous finite groups in computer science

Isomorphism testing is the area that comes to mind for me. In Graph Isomorphism (GI), the key techniques are Weisfeiler--Leman (WL) and Permutation Group Algorithms. At a high level, Babai uses WL to ...
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Wilson's proof of Iwasawa's Lemma

A point-stabilizer subgroup depends on a choice of point. However $\mathrm{Stab}(gx)=g\mathrm{Stab}(x)g^{-1}$ implies all points in an orbit have conjugate stabilizers. Thus, we can speak of "the&...
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6 votes

Does there exist a group homomorphism $(\mathbb{Q}_p, +) \to (\mathbb{R}, +)$?

As mentioned in the comments, as $\mathbb Q$-vector spaces (and hence also as abelian groups), $\mathbb Q_p\cong\mathbb R$. Thus, there are many group homomorphisms between them. This "morally&...
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2 votes
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Are the roots of unity the only algebraic subgroups of the multiplicative group?

Case of fields Let me begin by confirming your suspicion over a field. Claim (field case): Let $k$ be a field, and $H\subsetneq \mathbf{G}_{m,k}$. Then, $H$ is isomorphic to $\mu_{n,k}$ for some $n\...
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1 vote
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Does a free group $F$ of finite rank $n$ have finitely many retracts (as a subgroup)?

I close this question by giving the answer of @MoisheKohan in the comments: For $n\geq 2$, the statement is false because each free factor of $F_n$ is a retract and there are infinitely many free ...
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3 votes
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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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3 votes
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Are metacyclic $p$-groups semidirect products?

It turned out to be quite difficult to find a good link to answer this question. Therefore I will give here the full formulation of the theorem on metacyclic $p$-groups. Theorem. For odd $p$, every ...
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2 votes

Prove that $S$ is a coset of some subgroup of $G$ iff $S+S-S=S.$

This is @AtticusStonestrom's proof in more detail. Since $\varnothing\neq S$, let $t\in S$. It suffices to show that $H:=S-t=\{ s-t\mid s\in S\}$ is a subgroup of $G$, since then $S=H+t$ is a coset of ...
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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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1 vote
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Question about a proof for showing that $A_n$ has no subgroup of order $\frac{n!}{4}$ if $n>4$

Here $\sigma$ is an arbitrary three cycle. The proof shows that $\sigma \in H$. Thus each three cycle is in $H$. But $H$ cannot be $A_n$ since $H<A_n$ by definition.
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2 votes
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Can there be a different proof be given for: If $N$ is a subgroup of a group $G$ of index $2$, then $N$ is a normal subgroup of $G$

Since $[G:N]=2$, there is exactly two cosets of $N$ in $G$. One of them must be $eN=N.$ We have $$n\in N\iff nN=N.$$ (Can you prove this?) This is equivalent to $$a\notin N\iff aN\neq N.$$ If $b\neq a$...
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1 vote

If $G_1\cong G_2$, $H_1\triangleleft G_1$, $H_2 \triangleleft G_2$ and $G_1/H_1\cong G_2/H_2$, then is $H_1\cong H_2$?

The third condition is NOT correct. Consider $D_8$, the dihedral group of order 8, which represents the symmetries of a square in the plane. A presentation for $D_8$ is $$D_8 = \langle r, s ~|~ r^4=1, ...
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2 votes
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Quick question about a proof of the theorem: If $N$ is a subgroup of a group $G$ of index $2$, then $N$ is a normal subgroup of $G$

Let $a\in G$. The function $$\begin{align} \lambda_a:N&\to aN,\\ n&\mapsto an \end{align}$$ has inverse $$\begin{align} \lambda_{a^{-1}}:aN&\to N,\\ n&\mapsto a^{-1}n. \end{align}$$ It ...
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2 votes

Quick question about a proof of the theorem: If $N$ is a subgroup of a group $G$ of index $2$, then $N$ is a normal subgroup of $G$

There's a natural bijection between the elements of $N$ and the elements of $aN$ given by $\phi(g)= a\cdot g$ where $(\cdot)$ is the group multiplication. It's a good exercise to prove this map is ...
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1 vote
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Power of $ 2 $ congruent to 1 mod n

Ok so just summarizing what everyone else said, the solutions are exactly the multiples of $ o_n(2) $. So the number of solutions less than $ n $ is the floor function of $ n /o_2(n) $. When 2 is ...
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3 votes
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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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1 vote

First Isomorphism Theorem: Does each homomorphism has to be surjective? Is it possible to define an homomorphism $\phi:G\to H$ such that $|G|<|H|$?

Another standard homomorphism which is not surjective (for $|G|\ge 3$), is Cayley's one $a\mapsto(g\mapsto ag)$ from $G$ into $S_G$.
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2 votes
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First Isomorphism Theorem: Does each homomorphism has to be surjective? Is it possible to define an homomorphism $\phi:G\to H$ such that $|G|<|H|$?

Homomorphisms can be injective, surjective, isomorphic, or neither of them. It only has to be invariant on the structure: $\varphi(a\cdot b)=\varphi(a)\cdot \varphi(b)$ in case of (multiplicatively ...
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5 votes

Let $G$ be a group with $25$ elements and $E$ a $G$-set with $32$ elements. Show that there exists $a \in E$ such that $G_a=G$.

The orbits partition. Since the stabilisers orders have to divide $25$ (they're subgroups), they're all of order $1,5$ or $25$. So the same can be said for the orders of the orbits (orbit-stabilizer ...
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4 votes
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Group action for signal

For clarity, it should be made clear that $\mathcal{X}(\Omega)$ is not a signal, but the vector space of all signals $x\colon\Omega\to\mathcal{C}$ to some fixed space $\mathcal{C}$ which is suppressed ...
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4 votes
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Can we come up with a disjoint union of a subsets of the group $\Bbb{Z}$ such that they do not equal the cosets of a subgroup, yet they form a group?

It is not possible in any group. So, to be clear, we have a group $G$ and we are trying to come up wit a partition $G=\sqcup_i A_i$, where "termwise multiplication" yields to a group ...
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4 votes
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Why should I expect the generators of Lie Groups to be closed under the commutator?

So the proper formal answer should be that the vector fields on a Lie group have a natural Lie bracket (as do all vector fields on a general manifold). Then the Lie algebra can be identified with the ...
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2 votes

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

We have that the first group is isomorphic to $$\Bbb Z_4\times(\Bbb Z_3\times \Bbb Z_4)\times (\Bbb Z_5\times \Bbb Z_8),$$ while the second group is isomorphic to $$\Bbb Z_8\times(\Bbb Z_2\times \Bbb ...
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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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1 vote
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Is there a way to define other homorphisms, different from the conjugation mapping, in the definition of outer semidirect-products?

If $(A,+)$ is an abelian group then the only group conjugation is the identity, but $a\mapsto -a$ is an automorphism, and it is only the identity if every element of $A$ has $a+a=0.$ If $(A,+)$ is ...
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1 vote
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Writing a product of commensurable subgroups as a disjoint union

Let $g_1,\ldots,g_n$ be right coset representatives of $H_1 \cap gH_2g^{-1}$ in $gH_2g^{-1}$. Then every element $gh_2g^{-1}$ of $gH_2g^{-1}$ has the form $hg_i$ for some $h \in H_1 \cap gH_2g^{-1}$ ...
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Proving that a subgroup of a finitely generated abelian group is finitely generated

Here is a proof by induction on number of generators. For $n=1$ it is easy to see that subgroup of a cyclic group is always cyclic. Now assume that the statement is true for all abelian groups with ...
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1 vote

Definition of the Zappa–Szép product of groups in categorical terms

A definition by a universal property defines an object uniquely up to canonical isomorphism, so it cannot exist unless additional data is specified (the group is not defined by the subgroups ...
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2 votes
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Inverse of $3$ in multiplicative group $C_{20}.$

I am going to answer of your original question. Let $U(20) $ be the multiplicative group of $C_{20}$. We have $3\in U(20)$. What is the inverse of $3$ in $U(20) $? Suppose $3^{-1}=x$. Then $3x\equiv ...
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1 vote
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Express $\mathbb{Z}^2/B$ as a direct product of cyclic groups

Observe that $\Bbb Z\cong 2\Bbb Z\cong 3\Bbb Z$. Let $G=\langle a,b\mid ab=ba\rangle\cong \Bbb Z^2$. Suppose $$\begin{align} B&:=\langle a^2,b^2\rangle_G \\ &\cong 2\Bbb Z\times 2\Bbb Z\\ &...
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1 vote
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Point of confusions for: $H$ and $K$ are unequal subgroups of a group $G$, each of order $16$. Prove that $24 \leq |H \cup K| \leq 31$

It is given that order of $H$ and $K$ both are $16$. Again $H \cap K$ is a subgroup and as it is contained in both $H$ and $K$. So $|H \cap K|$ will divide $|H|$ and as well as $|K|$. So we have ...
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1 vote

Point of confusions for: $H$ and $K$ are unequal subgroups of a group $G$, each of order $16$. Prove that $24 \leq |H \cup K| \leq 31$

Since $H\ne K$, then $1\le\lvert H\cap K\rvert \le8\implies 24\le\lvert H\cup K\rvert \le31$. (BTW, $H\cup K$ won't be a group.)
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1 vote

Finite extensions of finitely generated free groups.

If $\Gamma$ is your example $\mathbb Z_2\rtimes_{\varphi(\bar 1)} F_2$ you are trying to extend $F_2$ by $\mathbb Z_2$ which means that you are trying to complete a short exact sequence $$0\to F_2\to\...
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1 vote

Do profinite completions commute with direct products?

Yes, profinite completion commutes with finite products. Rather than wondering what a subgroup of the product group looks like, it's easier to verify the universal property directly. The basic idea is ...
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1 vote
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How to Find Orbits and Stabilizers

The formula you are using is incorrect. It is not used for the whole set, but for one element only $$\forall\ s\in S,\lvert D_6\rvert=\lvert\text{Stab}(s)\rvert\cdot\lvert\text{Orb}(s)\rvert$$ That's ...
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