Second isomorphism theorem

Algebra Level 1

Let $S_3$ be the symmetric group on three letters. Let $H$ be the subgroup generated by the transposition $(12),$ and let $N$ be the subgroup generated by the transposition $(13).$

Then the second isomorphism theorem gives an isomorphism $HN/N \simeq H/(H \cap N),$ where $HN = \{ hn : h \in H, n \in N\}.$

Now $H \cap N = \{1\},$ so the right side has order $2.$ So the left side has order $2.$ Now $|N|=2$ and $|HN/N|=2,$ so $|HN|=4.$ But Lagrange's theorem says that $|HN|$ must divide $|S_3|,$ which is $6.$

What has gone wrong with this argument?

I. As defined above, $HN$ is not a subgroup of $S_3,$ so Lagrange's theorem does not apply.
II. The order of a quotient $G/K$ of finite groups is not always equal to $|G|/|K|.$
III. $N$ is not normal , so the second isomorphism theorem does not apply.

I and II II and III I, II, III I and III

2 solutions

Patrick Corn
Aug 2, 2016

The set $\{hn : h \in H, n \in N\}$ is only guaranteed to be a subgroup if $H$ or $N$ is normal. (Indeed, in this example, this set has four elements, so it is impossible for it to be a subgroup of $S_3.$ ) So statement I is true.

Statement II is false: it is always true that if $G$ is a finite group and $K$ is a normal subgroup, then $|G/K| = |G|/|K|.$

Statement III is correct: the Second Isomorphism Theorem only applies if $N$ is normal.

So the answer is I and III.

Shawn Ligocki
Sep 17, 2020

III. $N$ is not normal. We can see this by considering $g = (12)$ , then $gN = \{(12), (12)(13)\} = \{(12), (123)\}$ , but $Hg = \{(12), (13)(12)\} = \{(12), (132)\}$ . $(123) \ne (132)$ thus $N$ does not match the definition of a Normal Subgroup .

I. $HN$ is not a subgroup of $S_3$ . $HN = \{(), (12), (13), (12)(13)\} = \{(), (12), (13), (123)\}$ but $(123)^{-1} = (132) \notin HN$ therefore $HN$ is not a group at all.

Second isomorphism theorem

2 solutions

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