Tag Info

Hot answers tagged

6

It is correct, however, you might see the way similarly as: $$a(ab)b=a^2b^2=e=(ab)^2=a(ba)b.$$


5

Hints: Since $H$ is normal, what do you know about the conjugates of $H$? What do the Sylow theorems tell you about how different Sylow $p$-subgroups are related, for any given prime $p$?


4

Hints: Write the permutation as a composite of disjoint cycles, some of which could have infinite length. It is then enough to solve the problem for the individual cycles. Any cycle of length greater than $2$ is a composite of two elements of order $2$. For example, $(1,2,3,4,5) = f \circ g$ with $f=(2,5)(3,4)$, $g=(1,5)(2,4)$. This also applies to cycles ...


3

If $K$ is a normal subgroup of $G$ of order $4$, you may well argue in $H = G/K$, and reduce to show, via the correspondence theorem, that in a group of order $2 \cdot 3 \cdot 7 = 42$ there must be a normal subgroup of order $7$. To do this, just consider that the number of $7$-Sylow subgroups in $H$ must divide $42/7 = 6$, and be congruent to $1$ modulo ...


2

Hint: Instead of examining $\ker(f)$, take $\pi:H\rightarrow H/A$ to be the projection map and apply the same reasoning you give in the question to $\ker(\pi\circ f)$.


2

What @amWhy pointed is enough and I don't know if you are online to see my post or not. Other points that I can add: Your structure is closed under the given operation and indeed because of the nature of operation gcd, we have a*b=b*a: gcd(1,1)=1 gcd(1,2)=1 gcd(1,3)=1 gcd(1,4)=1 gcd(1,6)=1 gcd(1,12)=1 gcd(2,2)=2 gcd(2,3)=1 gcd(2,4)=2 ...


2

Another strategy: if $G$ is a finite group and $N \lhd G,$ then whenever $X$ is a subgroup of $G,$ then $XN = NX$ is a subgroup of $G$ of order $\frac{|X||N|}{|X \cap N|}.$ Now consider the case that $H = N$ and $X \in {\rm Syl}_{p}(G).$


1

It is easy to show that $e \in T$ (since $e \ast e = e$) and that $e$ acts as a unit element. Furthermore, $T$ contains inverses of its elements. By definition, the operation $\ast$ is associative. The most tricky part is showing that $T$ is closed under $\ast$. Suppose that $a$, $b \in T$. Then we have to show that $a \ast b$ is in $T$ as well, thus we have ...


1

First of all, your proof is absolutely fine! There's a more elegant way to prove this fact though. Since inverses are unique in groups, $(a^{-1})^{-1}$ is the unique element of the group satisfying $$ (a^{-1})^{-1} (a^{-1}) = (a^{-1}) (a^{-1})^{-1} = e. $$ On the other hand, we know that $a$ satisfies $$ a (a^{-1}) = (a^{-1}) a = e, $$ so we conclude ...


1

Hint: By the first isomorphism theorem, there exists a map $\phi: H\rightarrow H/N$ with kernel $N$. Consider $\phi^{-1}(K)=\{h\in H: \phi(h)\in K\}$.


1

More generally, for an arbitrary graph, you can consider all paths starting at a particular node; this "path space" has some nice properties. By restricting to LOOPS that start and end at some particular node, you get the loopspace of the graph; there's a natural kind of "multiplication" in which you traverse one loop and then the other; the composition ...


1

Your proof is definitely on the right track. The first part looks good. For the second part, there are just a few small issues. If you are using strong induction on $|G|$, then your base is for the $|G|=1$. You still need Cauchy's theorem later on, to mod out by a group of order $p$. Your argument breaks down in one case: suppose that $G=qn$, for $q$ a ...


1

Consider the composition $$ G\xrightarrow{f} H\xrightarrow{\pi}H/A. $$ Now $K:=\ker(\pi\circ f)\leq G$ is normal as the kernel of a group homomorphism, so either $K=G$, or $K=\{e\}$, since $G$ is simple. If $K=G$, then $f(G)\subseteq\ker\pi=A$, and you're done. If $K=\{e\}$, then $\pi\circ f$ is an injection of $G\hookrightarrow H/A$. But $[H:A]=2$, so ...


1

Hints: (a) If $\;H,K\;$ are subgroups of a group $\;G\;$ , then $\;HK:=\{hk\;:\; h\in H\,,\,k\in K\}\;$ is a subgroup of $\;G\;$ iff $\;HK=KH\;$ (b) If $\;H\;$ is a normal subgroup of $\;G\;$, then for any subgroup $\;K\;$ of $\;G\;$ we have that $\;HK=KH\;$ (c) In your question, you in fact only need to take any Sylow $\;7$-subgroup $\;K\;$ of $\;G\;$ to ...



Only top voted, non community-wiki answers of a minimum length are eligible