# Tag Info

1

Since $P \subseteq C_G(P) \subseteq N_G(P)$ and $P$ is a Sylow subgroup it follows that $|N_G(P):C_G(P)|$ is not divisible by $p$. So it must divide $p-1$. But $p$ is the smallest prime dividing the order $|G|$ and this means that $|N_G(P):C_G(P)|=1$.

1

Think about the first theorem of homorphism. It tells us that $G/ker(\phi)\cong Im(\phi)$ so if the group $G$ is finite so you' ll get the answer.

1

If $\phi :G\to F$ then $G/\operatorname{ker \phi}\cong \operatorname{Im\phi}=\phi(G)\implies o(G/\operatorname{ker \phi})=o(\operatorname{Im \phi})\implies \dfrac{o(G)}{o(\operatorname {Im \phi)}}=o(\operatorname{ker \phi})$ and $\operatorname{ker \phi}$ is a subgroup of $G$

0

They have to be disjoint because any $\sigma\in S_n$ can be written as a sequence of transpositions, these however are not disjoint. However if $\sigma,\tau\in S_n$ are not disjoint that means that they share one element, let's say $k$, such that in $\sigma$ we have $(k\to l)$ as a transposition, and in $\tau$ we have $(k\to r)$ which of course means that ...

7

Let $H$ be any quotient group of $\mathbb R$. Show for every $x\in H$, there is a $y\in H$ such that $y+y=x$. More generally, all quotient groups of any divisible group is a divisible group.

0

Suppose $G$ exists, let $p:R\rightarrow R/G=Z$ the projection, and $x\in R$ such that $p(x)=1$ for every integer $n>0$, $p(x)=1=np(x/n)$, impossible for $n>2$ since $p(x/n)\in Z$.

1

For the list of $n$'s which are included in the SmallGroups library, see : https://magma.maths.usyd.edu.au/magma/handbook/text/727 There are many missing n's which could be computed, even by hand (when $n=p^2q^2$, for example), but there are definitely some numbers less than $10000$ that are out of reach. For example, the number of groups of order $2048$ ...

0

$(2,6,16)$ does not seem right because $2$ is not a unit mod $4$, $6$ is not a unit mod $9$, and $16$ has order $2$ mod $17$. You need to find generators (aka primitive roots) mod $4$, mod $9$ and mod $17$. So try $(a,b,c)=(3,2,3)$ for instance.

1

The Universal Property of the free product $G*H$ of groups $G$ and $H$ is that there exist homomorphisms $i_G:G \to G*H$ and $i_H:H \to G*H$ such that, for any group $K$ and any homomorphisms $\tau_G:G \to K$ and $\tau_H:H \to K$, there is a unique homomorphism $\phi:G*H \to K$ with $\phi i_G=\tau_G$ and $\phi i_H=\tau_H$. It is not hard to prove uniqueness ...

0

Fun fact to know that it works also the other way around: let $G$ be abelian, if $\varphi(a)=a^n$ is an automorphism of $G$, then gcd$(n,|G|)=1$. Proof: assume that gcd$(n,|G|) \neq 1$ and let $p$ be a prime dividing both $n$ and $|G|$. By Cauchy's Theorem, there is an $a \neq 1$, such that $a^p = 1$. But then ...

1

Yes, this is a special case of confluence testing in the Knuth-Bendix completion process. It is described for polycyclic presentations (including for infinite groups) in Section 12.4 of "Handbook of Computational Group Theory" by Holt, Eick and O'Brien. For a polycyclic series of length $n$ there are about $n^3/6$ consistency conditions to check - most of ...

8

As other answers mention, it is not true that any Lie group is a matrix group; counterexamples include the universal cover of $SL_2(\mathbb{R})$ and the metaplectic group. However it is true that all compact Lie groups are matrix groups, as a consequence of the Peter-Weyl theorem. It is also true that every finite-dimensional Lie group has a ...

0

In general you can show a group $G$ of order $p^2$ ( with $p$ prime) is abelian, that is because $p$-groups have non-trivial center, so the center $Z(G)$ can have order $p$ or $p^2$ via Lagrange's theorem .But it also holds that if $\frac{G}{Z(G)}$ is cyclic then the group is abelian. So $Z(G)$ cannot be of order $3$ since otherwise $\frac{G}{Z(G)}$ would be ...

0

Just a hint : $a^{p} = \Big(b+ q p^{k}\Big)^p=\sum_{i=0}^p{p\choose i}\Big(qp^{k}\Big)^i \Big(b\Big)^{p-i}$ Then notice that for $i\ge 2$ you have $ik\ge k+1$, so $\sum_{i=2}^p{p\choose i}\Big(qp^{k}\Big)^i \Big(b\Big)^{p-i} \equiv 0 \ [p^{k+1}]$ That is why $a^p = b^p + qp^{k+1}b^{p-1} \equiv b^p \ [p^{k+1}]$

1

You are on the right way. You already proved that reflexivity of the equivalence relation implies that $e\in H$. Similarly other properties of equivalence relations imply that $H$ is a subgroup: Since $a\rho b\Leftrightarrow b\rho a$, we have $a\circ b^{-1} \in H \Leftrightarrow b \circ a^{-1} = (a \circ b^{-1})^{-1} \in H$, so $H$ is closed under taking ...

0

You are starting out good. Now, suppose that $a\in H$. Then you get $a\circ e^{-1}=a \in H$. By symmetry, $e\circ a^{-1}=a^{-1}\in H.$ Then all you have to do is closure under multiplication, which is transitivity.

0

It's a good first step, the next step is to check if $ab\in H$ and that inverses exist. These can be done at once by checking if $ab^{-1}\in H$ which we can easily do because if $a,b\in H$ then $a\rho b$ which means that $ab^{-1}\in H$ by definition of the relation, you're done.

0

to be normal we must have $xbx^{-1}\in H=\{e,b\}$, I use $e$ rather than $1$, for all $x$, is this the case? No because $$abb(ab)^{-1}=abbb^{-1}a^{-1}=aba^{-1}=b^{-1}a^{-2}=ba^{-2}$$ which is not in $H$

6

A Lie group is a group $(G,m,i)$ where $m \colon G \times G \rightarrow G$ is the multiplication and $i \colon G \rightarrow G$ is the inverse map that is also a smooth manifold such that $m$ and $i$ are smooth maps. Many Lie groups are subgroups of $\mathrm{GL}_n(\mathbb{R})$ but it is not true that any Lie group is isomorphic to a subgroup of ...

10

Not all Lie groups are matrix groups. Consider the metaplectic group. From wikipedia: The metaplectic group $M_{p_2}(\mathbb{R})$ is not a matrix group: it has no faithful finite-dimensional representations. Therefore, the question of its explicit realization is nontrivial. It has faithful irreducible infinite-dimensional representations, such as the ...

1

Hint Suppose $s + u$ were in $U$. Then, since $s + u \in U$ and $-u \in U$, and $U$ is a vector subspace... An analogous statement holds for more general groups $(G, \ast)$, if we replace the operation $+$ by the group operation $\ast$. Note that in the group setting, the operation is not in general commutative, and so we actually get one analog of the ...

1

It is fairly straight forward, a subspace must be closed under addition and scalar multiplication. If we assume that $s+u\in U$ then $(s+u)-u=s\in U$ which is a clear contradiction.

1

If $u+s$ would be in $U$, then $(u+s)-u=s$ would also be in $U$. A contradiction. This is also true for a group with the same proof.

0

$sgn(\sigma)=\prod_{i<j}\frac{\sigma(i)-\sigma(j)}{i-j}$ is quadratic, but straightforward to code.

3

You can call it Multiplication table, or Addition table if you're going by addition notion. In general you can just call it Cayley Table

1

Must all finite groups be cyclic? Certainly not. Just look at the Klein 4-group. To see a little more clearly why, just remember that a cyclic group of order $n$ must contain an element of order $n$, but as you see in the case of the Klein 4-group, all nonidentity elements have order $2$, and none have order $4$. All finite rings? (=Does the fact ...

0

You can either show that $\phi(x)=\phi(y)$ implies $x=y$ or if $x\neq y$ then $\phi(x)\neq\phi(y)$. Let's use the latter, if $x\neq y$ then we have that $xy^{-1}\neq e$. This gives us that $$\phi(xy^{-1})=\phi(x)\phi(y^{-1})=\phi(x)\phi(y)^{-1}=x^n (y^{n})^{-1}$$ for $x^n (y^{n})^{-1}=e$ we must have that $x^n=y^{-n}$ which is not possible as $x\neq y$

1

I assume you have shown that $\phi$ is a homomorphism. Showing that a homomorphism is injective is equivalent to showing that its kernel is trivial. Let $a \in \ker \phi$. We want to show that $a=1$. Let $g$ be the order of $G$ and let $k$ be the order of $a$. By Lagrange's theorem, $k \mid g$. Since $a \in \ker \phi$, we know that $a^n = 1$. Hence $k \mid ... 0 Hint: use the fact that$ \gcd(|G|, n ) = 1 $, hence, by Euclid's algorithm (or from the mentioned Bézout's identity), there exist integer$ x,y $such that $$x|G| + yn = 1$$ So $$a = a ^{x|G|}a^{yn} = a^{yn} = b^{yn} = b$$ 3 A finite group of order$n$will have exponent$n$if and only if its Sylow Subgroups are cyclic for every$p|n$. Suppose$p^r$is the highest power of$p$which divides$n$(and$r\ge 1)$. If the Sylow subgroups associated with$p$are not cyclic, then there is no element with order$p^r$, and hence no element with order divisible by$p^r$and the exponent ... 1 Using Chinese Remainder theorem you get : $$\mathbb{Z}/10\mathbb{Z}\times \mathbb{Z}/10\mathbb{Z}\text{ is isomorphic to }[\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}]\times[\mathbb{Z}/5\mathbb{Z}\times \mathbb{Z}/5\mathbb{Z}]$$ Now write : $$S_2:=\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$$ $$S_2:=\mathbb{Z}/5\mathbb{Z}\times ... 2 Third attempt. In general no. \mathbb Z_2 \times \mathbb Z_2 is the simplest exception. Many such examples will exist. I tried to generalize but I made mistakes. I think if n = pq for prime p and q. Then this will be true as some element g will have order p and only g^{kp} = 1 while another h will have order q and only h^{jq} = 1 so the only m ... 0 Here's the same argument, presented using additive notation (+) rather than multiplicative notation for the group operation. [This is conventional when a group is abelian, and indeed all cyclic groups are abelian.] Then instead of writing a^s = b^s for some integer s, we would instead say sa = sb, where the expression sa is an abbreviation for a ... 1 Let G_{s} and O_{s} be the stabilizer and orbit of an element s \in S. Then p^{r} = |G| = |G_{s}||O_{s}|, Hence |O_{s}| = p^{k}. On another hand we number the orbits that make up S arbitrarily, as O_{1}, ..., O_{k}. Then |S| = \sum_{i=1}^{k}|O_{i}| with |O_{i}| = p^{e_{i}}. If e_{i} \geq 1 for all i, then we get that p divide |S|, ... 1 HINT: You've already (correctly) concluded that E has to be 1; so really the only question is, "What are the possible values for C_3?" The properties a complex number z must satisfy, in order to be a candidate for C_3, are: z\not=1 (since C_3\not=E, and E represents 1); z^2\not=1; and z^3=1. Now there are two approaches for ... 5 If there are no such m and n, then the elements a,a^2,a^3,a^4,\dots are all distinct elements of R. But these elements then form an infinite subset of R, which contradicts the assumption that R was finite. (Note that this does not require R to be a domain or a to be nonzero.) 1 Pick a Sylow p-subgroup P and let G act on G/P by left multiplication: p\cdot gP = pgP. This gives a homomorphism G\to S(G/P)=S_r, which is clearly not identically equal to 1. Since G is simple, the kernel is thus trivial and hence S_r contains an isomorphic copy of G as a subgroup. So p^m r| r!. 0 assume you have two distinct subgroup H and K of order 5 ( otherwise the are the same). Now think about the set HK. 1 Let n_{5} denote the number of Sylow 5-subgroups. By Sylow Theorems, n_{5} \equiv 1 \mod 5 and n_{5}|4 and both conditions imply that n_{5} = 1. Hence the number of non-identity elements of order 5 is n_{5}(5-1) = 4. 2 It's not trivial, but the answer is no. See this MathOverflow question. In fact, a more general result of B.H. Neumann implies that a group can't be the union of finitely many cosets of infinite index subgroups. More precisely, it states that if a group is the union of n cosets of subgroups, then at least one of the subgroups must have index at most n. 2 The collection of all groups is very large. Saying it's "uncountable" is an understatement: for any infinite size (cardinality) of set in the mathematical universe, there is a group that large. No two groups of different sizes can even be isomorphic, much less identical. This collection is too big to form a set, given the systems of set theory that math is ... 0 Take G to be any group, let x,y be elements of G and observe that the commutator [x,y] = x'y'xy (here x' denotes the inverse of x in G) can be written as a product abc such that cba=1. Indeed, just take a=x', b=y'x and c=y. Now if x,y don't commute you get that abc is nontrivial, while cba is trivial and presto: you have infinitely many easy examples. ... 2 If \mathcal{X} and \mathcal{Y} are classes of groups that are closed under taking subgroups (such as abelian groups and finite groups: any subgroup of an abelian group is abelian, and any subgroup of a finite group is finite) then the class of \mathcal{X}-by-\mathcal{Y} groups (i.e., groups G with a normal subgroup N such that N is in ... 3 Correct. Alternatively stated: H is normal so it is a union of conjugacy classes; all those ccls have order dividing G, and hence have size either 1 or bigger than p (by definition of p as the smallest prime dividing |G|). Therefore it is a union of size-1 ccls. 0 Hint :Let H be such subgroup. Then H is divisible by 5 and by 3. Hence by 15. Now, if K is a subgroup of A_{5} of index m>1, then we get a non trivial homomorphism from A_{5} to S_{m}, and it is injective (why?)... then we get that m>4. Edited: Let K be a subgroup of A_{5} of index m>1. Then the action of A_{5} on the ... 1 D_6 contains normal subgroups G, H such that G \cong S_3 and H \cong \mathbb{Z}_2. Since G \cap H=\{1\} you can conclude D_6 \cong G \times H. I don't know if this helps, depends on your mathematical background. 2$$G=S_3\times\mathbb{Z_2}$$is not abelian and not cyclic hence it's not isomorphic to$\mathbb{Z_{12}}$and$\mathbb{Z_6}\times\mathbb{Z_2}$. Also the group$A_4$doesn't have an element of order$6$but$G$has, say$\{(1,2,3),1\}$, so correct option is$D_6$1 I'm not exactly sure how exactly you got rid of the squares in the exponents. That step definitely lacks justification and is probably incorrect. Here's what you can do: Let$A = \sum_{k=1}^{9}\xi_{19}^{k^2}$and$B = \sum_{k=1}^{9}\xi_{19}^{-k^2}$. Note that the exponents in$A$and$B$are precisely the sets of all squares and non-squares mod$19$. ... 0 If you know that$A_5$is simple and that all Sylow subgroups are conjugated, then this question is simple: since the set of all Sylow subgroups is invariant under conjugation, they generate a normal subgroup. So$K_{2,5}=A_5$has order$60$. In fact, this will work more generally for finite simple groups and any prime$p$dividing their order, for instance ... 1 Updated to reflect corrected question(but same idea): You can multiply together$16$of your equalities to get$ab^{16}a^{-1}=b^{32}\$. I don't know if that is the intended answer.

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