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Show that if $G$ is a group of order 6, then $G \cong \Bbb Z/6\Bbb Z$ or $G\cong S_3$

This is what I tried:
If there is an element $c$ of order 6, then $\langle c \rangle=G$. And we get that $G \cong \Bbb Z/6 \Bbb Z$. Assume there don't exist element of order 6.

From Cauchy I know that there exist an element $a$ with $|a|=2$, and $b$ with $b=3$. As the inverse of $b$ doesn't equal itself, I have now 4 distinct elements: $e,a,b,b^{-1}$. As there are no elements of order $6$, we have two options left. Option 1: $c$ with $|c|=3$, and $c^{-1}$ with $|c|=3$. Option 2: two different elements of order 2, $c,d$.

My intuition says that for option 1, $H= \{e,b,b^{-1},c,c^{-1} \}$ would be a subgroup of order $5$, which would give a contradiction, but I don't know if this is true/how I could prove this.

I also don't know how I could prove that option 2 must be $S_3$. I think that $G= \{e,a,b,b^{-1},c,d \}$. Looks very similar to $D_3$. But I don't know how I could prove this rigoursly.

Can anybody help me finish my proof ? If there are other ways to see this, I'd be glad to hear them as well!

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marked as duplicate by Derek Holt, MJD, Mikko Korhonen, user1729, Davide Giraudo Apr 16 '13 at 14:18

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

If $a$ has order 2 and $b$ has order 3 and commute, then $ab$ has order 6 and $G$ is cyclic. If not, $1$, $a$, $b$, $ab$, $ba$ and $b^2=b^{-1}$ are 6 elements. Do they form a group? – Andrea Mori Apr 16 '13 at 12:42
Show that $G=\{{\,1,a,b,ab,b^2,ab^2\,\}}$. Show that either $ab=ba$ or $ab=b^2a$. Show that the 1st option leads to the cyclic group, the 2nd, the symmetric/dihedral. – Gerry Myerson Apr 16 '13 at 12:47

Instead of introducing a new element called $c$, we can use the group structure to show that the elements $ab$ and $ab^2$ are the final two elements of the group, that is $G=\{e,a,b,b^2,ab,ab^2\}$. Notice that if $ab$ were equal to any of $e,a,b$, or $b^{-1}=b^2$, we would arrive at the contradictions $a=b^{-1}$, $b=e$, $a=e$, and $a=b$ respectively. Similarly, see if you can show that $ab^2$ must be distinct from the elements $e,a,b,b^2,$ and $ab$.

In a manner similar to the one above, we can now show that the element $ba$ (notice the order) can only be equal to either $ab$ or $ab^2$ in our list of elements of $G$ without arriving at some contradiction. If $ba=ab$, then $|ab|=6$, so that $G$ is cyclic. If $ba=ab^2$, see if you can write down a group isomorphism between $G$ and $S_3$, and show that the images of the elements of $G$ in $S_3$ satisfy the same relations their pre-images do in $G$.

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