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I am working on the following group-theory exercises but I'm a little confused by how to begin proving them.

Let $G$ be a group. Call $g \in G$ a $torsion \ element$ if $g$ has finite order $g^k = e$ for some $k \neq 0$) and write $G_{tors}$ for the set of torsion elements. Say that $g$ is $p-power \ torsion$ if its order is a power of $p$. For an abelian group write $A[p^{\infty}]$ for the set of its $p-$power torsion elements.

$\bf{Question \ \#1:}$ Fix an abelian group $A$.

(a) Show that $A_{tors} $ and $A[p^{\infty}]$ are subgroups of $A$.

(b) Show that $A[p^{\infty}]$ is a p-Sylow subgroup of $A$. Take it as given that if $A$ is finite, $A = \Pi_p A[p^{\infty}]$ as an internal direct product.

(c) Show that $A/A_{tor}$ is $torsion-free$: $(A/A_{tors})_{tors} = \{e\}$.

$\bf{Question \ \#2:}$ Find the Sylow subgroups of $C_{360} \times C_{300} \times C_{200} \times C_{150}$.


For (a), can I just show that if $a,b \in A_{tors} $ and $a,b \in A[p^{\infty}]$ then $ab^{-1} \in A_{tors}$ and $ab^{-1} \in A[p^{\infty}]$?

For (b), can I also just do the same and show that $a,b \in A[p^{ \infty}]$ implies that $ab^{-1} \in A[p^{ \infty}]$?

For (c), I am not sure I even understand the question. Is a group torsion-free if all of its elements are of infinite order?

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  • $\begingroup$ For $(c)$, yes, this is the question. Show that $A/A_{tor}$ has no elements of finite order, except $1$ (which has order $1$) - see here. $\endgroup$ – Dietrich Burde Nov 24 '15 at 21:02
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For (a), can I just show that if $a,b \in A_{tors} $ and $a,b \in A[p^{\infty}]$ then $ab^{-1} \in A_{tors}$ and $ab^{-1} \in A[p^{\infty}]$?

Yes, for a non-empty subset $H \subseteq G$ it holds that $H$ is a subgroup of $G$ iff for all $a, b \in H$ we have $a b^{-1} \in H$. In this case the non-empty subset bit is trivial (but deserves mentioning nonetheless).

For (b), can I also just do the same and show that $a,b \in A[p^{ \infty}]$ implies that $ab^{-1} \in A[p^{ \infty}]$?

No, this just proves it is a subgroup. You need to prove that

  1. it is a $p$-group (i.e. all elements have orders a power of $p$) – trivial in this case
  2. it is maximal among all other $p$-groups, that is if $H$ is any other $p$-group that we have $H \subseteq A[p^\infty]$.

For (c), I am not sure I even understand the question. Is a group torsion-free if all of its elements are of infinite order?

Yes, that's correct.

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