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9

Hint: Here is a simple proof idea that every group of even order must have an element of order $2$. Pair every element in $G \backslash \{ e \}$ with its inverse. If all pairs consist of two different elements then $G \backslash \{ e \}$ would have an even number of elements. What does it mean that $a=a^{-1}$?

8

Hint: Let $f:H\to G$ be an isomorphism and let $K\subseteq H$ be a normal subgroup. What can you say about $f(K)\subseteq G$?

6

Approach without Sylow's theorem: By what you've shown, all you need to do is discount the possibility that all group elements have order $3$ or $1$. The only element with order $1$ is the identity. What can you say about a group that consists of the identity, and $11$ elements of order $3$? Hint: the elements of order $3$ can be partitioned into pairs $\{g,... 3 The set you're talking about -$\mathbb{Z}/8\mathbb{Z}-\{0\}$- consists of everything in$\mathbb{Z}/8\mathbb{Z}$except for$0$(or rather,$[0]$- it's a bit sloppy notation). That's what the "$-$" means. So$2$and$4$(or rather,$[2]$and$[4]$) are in the set, but their product is not. 3 Hint Let$V=\{1,(12)(34),(13)(24),(14)(23)\}$a subgroup of$\mathfrak S_4$. This group has 4 element, and is normal in$\mathfrak S_4$. If you can prove that $$\mathfrak S_4/V\cong \mathfrak S_3,$$ then, if $$\varphi:\mathfrak S_4/V\longrightarrow \mathfrak S_3$$ is such an isomorphism, then,$\varphi\circ \pi$is the researched homomorphism where $$\pi:\... 3 |G| refers to the order of G (abelian or not). Action by left multiplication is always transitive (consequence of the so-called 'sudoku property' of groups). The conjugation action however, is not transitive for any non-trivial group G. 2 Hint: Consider the Sylow 2-subgroups of G, which have order 4. Hope this helps. 2 I found this example in "Lectures on Finitely Generated Solvable Groups" By Katalin A. Bencsath, Marianna C. Bonanome, Margaret H. Dean, Marcos Zyman. Let$$A= \begin{pmatrix} 1 & 0 & 0 \\ 0 & x & 0 \\ 0 & 0 & 1 \end{pmatrix}, B=\begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{pmatrix}$$where these ... 2 Recall a group is simple if it doesn't have a normal subgroup other than itself and the identity group, or put another way "no non trivial normal subgroup". So a natural proof by contradiction arises. Since$G \cong H$we can consider an isomorphism$\pi: H \rightarrow G$. Suppose now that$H$has a non trivial normal subgroup$K$. we will show that$\pi(...

2

$\mathbb Z_2$ is abelian. Thus conjugation can't change the first member of an element of $\mathbb Z_2\oplus S_3$.

2

Yes, this is correct. Another way of referring to the kernel of homomorphism is the unit circle, because all of the complex numbers in this kernel create a circle of radius $1$ around $0$ on the complex plane. Good job!

2

A finite group of order $p$ or $p^2$ for some prime $p$ is abelian. Then the proper subgroups of groups of order $1$ up to $11$ are always abelian. For $12$, we have $S_3\times C_2$ which has a non-abelian subgroup ($S_3$).

1

There seems to be confusion about subgroups direct products, or what direct sums are. You claim: ...for each subgroup $H\oplus K$ of $G_1\oplus G_2$, the groups $H$ and $K$ are subgroups of $G_1$ and $G_2$. But here $H$ and $K$ are supposed to be subgroups of $G_1$ and $G_2$ to begin with; how else can $H\oplus K$ be considered a subgroup of $G_1\oplus ... 1 Thanks to the hint of fulges in comments, I worked out the following proof: Remark: here$\subset$means strict inclusion. Let$G_0 \subset ... \subset G$be a normal series of maximal length$n$in$G$. If$G_0 \neq \{e \}$, then we can construct another normal series$\{ e \} \subset G_0 \subset ... \subset G$having a greater length$n+1 > n$. So$...

1

Your first reformulation of the problem is not correct; it suffices to show that one of the quotients is not simple, not all of them. It is true that $K/H$ is normal in $P/H$ if and only if $K$ is normal in $P$. But you don't need this fact because all groups concerned are abelian, so all subgroups in all quotients are normal. The last statement you wish ...

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