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From: kyokoyoshida123@gmail.com
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Date: Wed, 6 May 2009 14:25:53 -0700 (PDT)
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いつも大変お世話になっております。

Let H be a normal subgroup of a group G(i.e., for all g∈G;gH=Hg), and
let m:=[G:H]. Show that a^m∈H for every a∈G.(Hint: Look at aH)

という問題なのですがLagrangeの定理からm|#Gが言えると思います。
あと,a^mH=a^{m-1}aH=a^{m-1}Ha=…=Ha^mとなると思いますがこれからどうすればいいのでしょうか?

吉田京子