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5

While the other answers lead you to the precise order of $GL_n(\mathbf{F})$, I'll explain a simple way to see the inequality you asked about. As you note, any $n \times n$ matrix has $n^2$ entries; if these entries are coming from $\mathbf{F}$, then the set of all $n \times n$ matrices has order $q^{n^2}$, since there are $q$ choices for each of the $n^2$ ...

4

No, this relies heavily on the fact that $n=1$. Here is a general proof. Suppose $P\leq G$ is a Sylow $2$-subgroup. Then there is a homomorphism $\pi:G\to S_{m}$ induced by the action of $G$ on the set of left cosets of $P$. If $G$ is simple then $\pi$ must be injective because it is not trivial. This means that $$2^nm\leq m!$$ so $$2^n\leq (m-1)!$$ But ...

3

The group $S_3$ has only $\{\mathit{Id}\}$, $A_3$ (the alternating subgroup) and $S_3$ as normal subgroups. If $\ker\varphi=S_3$, then we have the trivial homomorphism. We can rule out $\ker\varphi=\{\mathit{Id}\}$, because $S_3$ is not abelian. Let's assume $\ker\varphi=A_3$. Then $\varphi$ induces an injective homomorphism $\hat\varphi\colon ... 3 Hint: Every 3-cycle must map to some$a\in\mathbb Z_4$with the property$a+a+a=0$, and there is only one possibility for that. Since the product of two different transpositions is a 3-cycle, this leaves very few possibilities for what the image of each of the three transpositions can be. 2 No. This cannot be done with the cyclic group of order$3$. Let$G = \mathbb{Z}/3$with generator$1$, and consider the effect of any one step of the given process. If the probability distribution before the step is$(P_0,P_1,P_2)$and the step adds$1$with probability$p$, then the probability distribution after the step is $$(P_0',P_1',P_2') \;=\; ... 2 Since we know that every group of order n is isomorphic to a subgroup of S_n. So suppose \phi:G\to S_{2m} is that homomorphism. Now since |G|=2m~(even) hence \exists an element a whoes order is 2. Define a_l:G\to G as$$a_l (x)=ax$$then clearly$a_l$is a bijection so$a_l\in sym(G)=S_{2m}$Notice that$a_l$gives a transposition with ... 2 Taken from http://science.kennesaw.edu/~sellerme/sfehtml/classes/math4361/chapter4section1outline.pdf Let$G=\{1,2,3,4,5\}$with multiplication table \begin{array}{|c||c|c|c|c|c|} \hline *&1&2&3&4&5\\ \hline \hline 1&1&2&3&4&5\\ \hline 2&2&1&4&5&3\\ \hline 3&3&4&5&2&1\\ \hline ... 2 A minimal normal subgroup of any finite group is a direct product of isomorphic simple groups. So if$G$is solvable then these must be abelian simple groups i.e. cyclic of order$p$for some prime$p$. So$N$is a$p$-group. 1 This is not an answer, but rather an extended comment. In your comments, you state that the identity element$s$is unique. I'm not sure how to prove this from the axioms, but OK. Anyway, I'm not convinced you've looked at enough examples. Why don't you do$D_8$and$Q_8$? If you take a group$G$, with$u_1$the usual group multiplication, then your ... 1 No to both questions. For example for the first, in the generalized quaternion groups, there is a unique element of order$2$, which is thus contained in any non-trivial subgroup. For the second, consider either of the groups the group$S_3$or$S_4$which are solvable but have trivial centers. 1 Clearly the factor group you give is abelian and thus the direct product of cyclic groups and Derek Holt's comment gives you that you do not need more than$d$. If$d$is also the minimal number of generators, Burnsides basis theorem shows that for$M=[G,G]G^p$the group$G/M$is the direct product of$d$cyclic groups of order$p$. Since your$N\le M$, ... 1 Like anon I assume that the question is about the centralizer of$P$(with that typo/mistranslation the question makes IMHO more sense). The highest power of$p$that divides$p!$is$p^1$. Therefore the Sylow subgroups$P$are all cyclic of order$p$. A permutation of prime order must be a product disjoint$p$-cycles (ignoring fixed points). In the case of ... 1 All you have done is correct. However it seems you made some claims which at first sight are correct, but in fact they could be based on some confusion. One of them is the following: when$G=\mathbb Z/p^{\alpha}\mathbb Z\times\mathbb Z/p^{\beta}\mathbb Z$you said that$\mathbb Z/p\mathbb Z\subset G$. This is right up to isomorphism: we identify$\mathbb ...

1

I'm not sure what exactly I'm allowed to use or what Rotman had in mind, but I think this approach uses no more group theory than the (unique) factorization of permutations and the basic idea of what a homomorphism is. If $n$ is odd there exists a subset $S\subset S_n$ of pairwise commuting elements of order $2$ with $|S|=\tfrac{n-1}{2}$. For example ...

1

As Taylor said, just find a generator (and it's the obvious one. If still confused, I can give a hint). In general, you can show that $\mathbb{Z}_m\times\mathbb{Z}_n\cong\mathbb{Z}_{mn}\iff \gcd(m,n)=1$

1

I am going to start where Derek finished, with the aim of giving some justifications as to why the groups are mutually non-isomorphic. To summarize, there are seven possibilities: we may choose $x, y, z$ such that $P = \langle x \rangle \cong \mathbf{Z}_5$, $Q = \langle y \rangle \cong \mathbf{Z}_{11}$, $R = \langle z \rangle \cong \mathbf{Z}_{61}$, and ...

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