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I'm working through Judson's Abstract Algebra: Theory and Applications and I came across this lemma (pg 71 in the text, 83 in the pdf):

Lemma 6.3. Let H be a subgroup of a group G and suppose that $g_1, g_2 \in G$. The following conditions are equivalent.

  1. $g_1 H = g_2 H$;
  2. $H g_1 ^{-1} = H g_2 ^{-1}$;
  3. $g_1 H \subseteq g_2 H$;
  4. $g_1 \in g_2H$;
  5. $g_1 ^{-1} g_2 \in H$.

I understand why all properties hold except 2. Intuitively, it seems like 2 would hold, but I don't see why it must be the case. I also only believe 5 if I assume 2.

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  • 6
    $\begingroup$ HINT: If $g_1H=g_2H$, then $g_1=g_2h$ for some $h\in H$. (Number 4. is sort of the key to everything.) $\endgroup$ – Ted Shifrin Dec 6 '16 at 4:37
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$H$ is a subgroup so $H^{-1}=H$ (is this clear? think this on subgroup level but not elementwise although it is true; then you can immediately conclude how $(1) \Longleftrightarrow (2)$ is obvious!)

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  • $\begingroup$ You may know well that $(ab)^{-1}=b^{-1}a^{-1}$! $\endgroup$ – p Groups Dec 6 '16 at 4:47
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(1). $A^{-1}=\{a^{-1}:a\in A\}.$

(2). If $H$ is a group then $H^{-1}=H.$

(3). $A=B\iff A^{-1}=B^{-1}.$

(4). $(xA)^{-1}=A^{-1}x^{-1}.$

(5). Therefore if $H$ is a group then $g_1H =g_2H\iff (g_1H)^{-1}=(g_2H)^{-1}\iff$ $\iff H^{-1}g_1^{-1}=H^{-1}g_2^{-1}\iff Hg^{-1}=Hg_2^{-1}.$

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