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<P>ÀÌÇö³à ¼±»ý´ÔÀÇ ´äº¯ÀÔ´Ï´Ù.</P> <P>&nbsp;</P> <P>¾î¶² »ç¶÷ÀÌ ÀÌ·¸°Ô Áú¹®À» Çß½À´Ï´Ù. <BR>I would like to know following solution of S_3 is not decomposible <BR>is correct or not ? <BR><BR>Áú¹®ÀÚÀÇ »ý°¢ÀÔ´Ï´Ù. <BR><BR>sol&gt; <BR>Since |S_3| = 6 , <BR>If S_3 is isomorphic to H x K (H,K are subgroup of G) and H=/={1}=/=K, <BR>then W.L.O.G it must be H=&lt;(12)&gt;, K=&lt;(123)&gt; or H=&lt;(23)&gt;,K=&lt;(123)&gt; or <BR>H=&lt;(13)&gt;, K=&lt;(123)&gt;. <BR>Now, in all above three cases, <BR>since H x K ~= C_6 (cyclic group of order 6) , <BR>H x K is cyclic group but S_3 is not a cyclic group. <BR>therefore, S_3 has no subgroups H,K satisfying S_3 ~= H x K, <BR>H=/={1}=/=K. <BR>=&gt; so, S_3 is not decomposible. <BR><BR>I read this solution at my book but, I think this solution is false <BR>because above H is not a normal subgroup of G. <BR><BR>I know the definition of decomposible group as follows : <BR>Any group G is decomposible iff there are normal subgroups H,K of G <BR>satisfying <BR>G~= H x K , |H|=/=1=/=|K| <BR><BR>¸®ÇÃÀÌ ÀÌ·¸°Ô ´Þ·È½À´Ï´Ù. <BR>That would be another way of seeing that G is not indecomposable (it <BR>would have to have a normal subgroup of order 2). But there is nothing <BR>wrong with the argument given: the product of a cyclic group of order <BR>2 and one of order 3 is necessarily cyclic of order 6. Since G is not cyclic, that s it. <BR><BR>±×·¯³ª »ç½Ç Àü ¸ð¸£°Ú½À´Ï´Ù. Çö´ë´ë¼ö¸¦ Á¢ÇÑÁö ¿À·¡µÇ¼­.. <BR>´Ù¸¸ ÀÎÅͳݿ¡ ¶°µµ´Â Áú¹®°ú ´äÀÌ Àֱ淡 ¿Ã¸³´Ï´Ù. ½º½º·Î ÆÇ´ÜÇϼſä..</P>