# Tag Info

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By Sylow's theorems, the group must have a normal $7$-subgroup and a normal $13$-subgroup, and so it is the direct product of these two subgroups, and is cyclic. Then the problem then becomes a special case of this problem. The upshot is that the only possible size of $G/Z(G)$, if it is cyclic, is $1$.

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$G = D_{10} := \left < \{r. s| r^5 = 1, s^2 = 1, rs = sr^{-1} \} \right>$ be the dihedral group of order $10$ with the standard generators $r, s.$ Let $H := \left< s \right>.$ Then $[G : H ] = 5.$ But $H$ is not normal in $G$, since $rsr^{-1} = r^2s = sr^{-2} \neq s, e.$ [Here $p = 5, m = 2.$]

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The following idea is super-simple. However, it only works if $n$ is odd (so it works precisely half the time!). But as it is so simple I thought it would be nice to record it here. Suppose $n$ is odd. Note that $S_n$ can be generated by the elements $\alpha:=(1, 2)$ and $\beta:=(1, 2, \ldots, n)$. Now, every subgroup $H$ of index 2 is the kernel of some ...

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$\mathbb{Z}_{p^2}$ has a unique subgroup of order $p$, and that subgroup is the kernel of the homomorphism $g$.

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Since the question only asks for composition factors, you don't really need to identify what subgroups $X_i$ you'd like to use. Instead you just need to describe composition factors for (any, pick your favorite) cyclic group of order p-1. Then, there is nothing special about p-1 being the product of two primes -- the composition factors are simply the ...

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$G/Z(G)$ can not be cyclic if $G$ is nonabelian and the group of order $91$ is cyclic as $7$ does not divide $13-1$.

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Sylow subgroups of $G\times H$ are also products of Sylow subgroups of $G$ and of $H$. Now what are the Sylow subgroups of $S_3$?

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Quang's method is undoubtedly the easier approach here, although there are alternative ways of attacking this problem. Sylow $2$-subgroups have order $4$ in $S_3 \times S_3$, however, none of them can be isomorphic to $\mathbb{Z}_4$. Why? This would imply there exists an element of order $4$ in $S_3 \times S_3$. Can you see why there cannot exist such ...

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$L(C)$ is solvable, the other two aren't. ${\rm PSL}(2,\mathbb{C})$ contains a subgroup isomorphic to $\mathbb{R}^2$, ${\rm PSL}(2,\mathbb{R})$ doesn't.

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An inner automorphism will preserve the spectrum of matrices. If you look at some diagonal matrix like $$\mathrm{Diag}(2,2,\dots,2,2^{-(n-1)})$$ then the transpose of its inverse has a different spectrum, at least for $n>2$. For the $n=2$ case I'm not immediately sure how to proceed. EDIT As @JyrkiLahtonen points out, the statement is false for $n=2$. ...

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Yes, it is true. Consider the assertion that $[x,y_1 \cdots y_n]$ can be expressed as a product of conjugates of $[x,y_n],\ldots,[x,y_1]$, in that order. For $n=2$, it can be verified that $[x,y_1 y_2] = [x,y_2][x,y_1]^{y_2}$. Fix $n$ and assume the assertion holds for smaller values of $n$. Then $[x,y_1 \cdots y_n] = [x,y_n] [x,y_1 \cdots y_{n-1}]^{y_n}$. ...

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