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### If $m\mid n$ can we find a group $G$ of order $n$ with a subgroup $H$ of order $m$, i.e. can every divisibility be proved group theoretically?

More precisely, if $m\mid n$, can we always find a group $G$ of order $n$ and a subgroup $H$ of $G$ of order $m$? Yes. For each $n$, there exists a cyclic group of order $n$. In the cyclic group of ...
• 45.1k
Accepted

1 vote
Accepted

### $g$ lies in the centre of $G$ iff $\vert\chi(g)\vert=\vert\chi(1)\vert$ for all irreducible characters $\chi$

Let $\rho$ be the $n$-dimensional irreducible representation with character $\chi$. If $|\chi(g)|=|\chi(1)|=n$, since $\rho(g)^m=\rho(g^m)=\rho(1)=1$ for some $m$, the eigenvalues of $\rho(g)$ are $n$ ...
• 15.1k
1 vote

### Given a tuple of $k$ distinct integers, is there a generator list in a $\mathbb{Z}/n\mathbb{Z}$ that matches the tuple?

Basically, you want geometric sequence modulo some $n$, so $q_i^2 \equiv q_{i-1}q_{i+1}\pmod n$, i.e. $n\mid q_i^2-q_{i-1}q_{i+1}$, so to find counterexample you can take any triple $(q_1,q_2,q_3)$ ...
• 23.1k
1 vote

### How many homomorphisms are there from $D_5$ to $V_4$?

The commutator of $D_5$ is $\langle r\rangle \cong \Bbb Z_5.$ As a result it's in the kernel (there's two ways to see that: $V_4$ has order $4,$ and it's abelian) , and the homomorphisms have every ...
• 8,512
1 vote

Since $2\nmid 5$, all the four elements of order $5$ must be sent to $1$. As for the five elements of order $2$, $\varphi(sr^k)=$ $\varphi(s)\varphi(r^k)=$ $\varphi(s)$. Therefore, for $V_4=\{1,a,b,ab\... • 3,035 1 vote ### Group of order$p^{\alpha}q$is not simple. Since the Sylow$p$-subgroups are all pairwise conjugate (Sylow II), their intersection is the normal core in$G\$ of any of them. So, once you know this intersection is not trivial, you get a ...
• 3,035

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