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Clash Royale CLAN TAG #URR8PPP up vote 1 down vote favorite I want to use induction to show that the sum $x_1 + dots + x_n$ of real numbers is defined independently of parentheses to specify order of addition. I know how to apply induction(base, assumption, k+1 applying inductive hypothesis). Here I am not sure what the base would be. I have two ideas: 1) First case is $(x_1 + x_2)+x_3+dots+x_n$ and work through to $x_1+x_2+dots+x_n-2 + (x_n-1 + x_n)$ 2) Start with $(x_1+x_2)+x_3=x_1+(x_2+x_3)$ and work up in number of elements to the full case. Both seem wrong, I have no idea what to actually do. I imagine above is sufficient effort, although I have shown no working. Before you downvote, please tell me why you are planning it, and I will edit. induction share | cite | improve this question edited Jan 29 '15 at 7:12 Asaf Karagila ♦ 295k 32 411 738 asked Jan 29 '15 at 5:47 qqqqq 6 1 You cool with Peano's ax...

Clash Royale CLAN TAG #URR8PPP up vote 1 down vote favorite Let $G$ be a finite group and $p$ a prime number. Recall that a subgroup $H$ of $G$ is strongly $p$-embedded if $p | |H|,$ and for each $x in G setminus H,$ $H cap ^xH$ has order prime to $p,$ where $^xH$ denotes the group we get after conjugation by $x.$ I am reading a proof which claims that any strongly $p$-embedded group contains a $p$-Sylow group of $G$. I can prove this fact by other means, but I would like to understand the proof I am reading. This is the argument I am reading: Assume $H$ is strongly $p$-embedded. Since $p | |H|,$ $p$ contains an element $g$ of order $p.$ Choose any $S in Syl_p(G)$ such that $g in S.$ For $x in Z(S)$ of order $p,$ $$g in H cap ^xH$$ implies $x in H.$ Hence for all $y in S,$ $$x in H cap ^yH $$ implies that $y in H.$ Thus $S subset H.$ Here is what I do not understand with the proof: Why does it matter that $x in Z(S)$ has order $p?$ It seems to me that for any $x in Z(S),$ ...