# Tag Info

8

By Schur's theorem (well, one of the things called that, at least), $G'$ is finite if $Z(G)$ has finite index. Since $G$ contains no torsion, $G'$ must vanish; that is, $G$ is abelian.

5

Yes. Suppose $G\neq 1$. Since there exists an element with prime order, and all non-identity elements of $G$ have the same order, $G$ is a $p$-group. Therefore $Z(G)\neq 1$. But automorphisms preserve the center, so $Z(G)=G$.

5

For me, you statement is false: $\mathbb{R}$ is a $\mathbb{Q}$-vector space of infinite dimension, so two proper subspaces $A,B$ may be found so that $\mathbb{R}= A \oplus B$. In particular, it gives a decomposition of $\mathbb{R}$ as an additive group.

2

Suppose that $H$ is that unique minimal subgroup of $G$. Then for any $g\in G$: $H\subset \langle g\rangle$ implying $H$ is cyclic, Since $H$ is minimal $|H|=p$ is a prime, $\langle g\rangle$ has order $p^n$ for some $n$, $H=\langle g^{p^{n-1}}\rangle$ Now suppose that $h$ is an element of $G$ with the maximal order $ord(h)=p^m$, then $H=\langle ... 2$C'$is not a direct summand in$M$: if there is$C''$such that$C'+C''=M$and$C'\cap C''=0$, then$|C''|=4$, so$C''$is either cyclic or Klein. Since the elements of order four are$(2,0)$,$(2,1)$,$(6,0)$, respectively$(6,1)$it's clear that$C''=\langle(2,0)\rangle=\langle(6,0)\rangle$. But$|C'+C''|=8$, so$C'+C''\ne M$. When$C''$is of Klein type, ... 1 Let$R$be a ring and$M$an$R$-module. Suppose$u,v\in R^\times$are units. Then for any$\pi\in R$, $$\frac{M}{u\pi\color{Green}{vM}}=\frac{M}{u\pi\color{Green}{M}}\cong \frac{\color{Blue}{u^{-1}M}}{\pi M}=\frac{\color{Blue}{M}}{\pi M} \implies \frac{M}{u\pi v M}\cong\frac{M}{\pi M}.$$ The isomorphism is given by multiplication-by-$u^{-1}$. Let$M=\Bbb ...

1

What you’ve written so far seems good to me, (modulo the now-fixed mix-up). To go on: Try to find a group arrow $G → F$. Use the universal property of the free group on $X$ (of which $G$ is a factor group). Then use the first isomorphism theorem (or the universal property of factor groups, whatever you call it). Then don’t show that $f$ is monic, but that ...

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