Is the following question I found in a past exam paper a trick question ?

$$G=\Bbb Z_{120}, M=\{0,10,20,...,110\}, N=\{0,4,8,12,..,116\}.$$ Then identify the subgroup of $G$ given by $MN$.

The reason I think it's a trick question is that $|M|=11,|N|=29$ so wouldn't this imply that $|MN|=319$ and then by Lagrange can't be a subgroup as the order of this group doesn't divide G ?


It's more customary to write $M+N$ when the group is written with additive notation.

First you're wrong in $|M|$ and $|N|$, which are $12$ and $30$ respectively. On the other hand, $|M+N|$ cannot be $360$, because you only have $120$ elements to choose from.

The general formula, holding also for nonabelian groups, is $$ |M+N|=\frac{|M|\,|N|}{|M\cap N|} $$ The formula is easier to prove for abelian groups. Still using additive notation, consider the surjective group homomorphism $$ \sigma\colon M\times N\to M+N,\qquad (x,y)\mapsto x-y $$ Its kernel is the set of pairs $(x,y)$ such that $x=y$, which is in an obvious bijection with $M\cap N$. The homomorphism theorem says that $$ (M\times N)/\!\ker\sigma\cong M+N $$ so that $$ |M+N|=\frac{|M\times N|}{\lvert\ker\sigma\rvert}=\frac{|M|\,|N|}{|M\cap N|} $$

Can you tell what's $M\cap N$?

A different way is to notice that $M=10\mathbb{Z}/120\mathbb{Z}$ and $N=4\mathbb{Z}/120\mathbb{Z}$, so $$ M+N=(10\mathbb{Z}+4\mathbb{Z})/120\mathbb{Z}=2\mathbb{Z}/120\mathbb{Z} $$ because $2=\gcd(10,4)$.

  • $\begingroup$ $M\cap N$={0,20,40,60,80,100} , I think which would mean by your formula that |M+N|=360/6=60 which does divide 120 and so can be a subgroup. also looking at the part of your answer ( which by the way is fantastic , thanks for that :)) where you say $M+N=(10\mathbb{Z}+4\mathbb{Z})/120\mathbb{Z}=2\mathbb{Z}/120\mathbb{Z}$ does that mean that $M+N =2 \Bbb Z/120 \Bbb Z \cong \Bbb Z_{60}$ ? $\endgroup$ – excalibirr May 13 '18 at 23:18
  • 2
    $\begingroup$ @exodius Excellent. $\endgroup$ – egreg May 13 '18 at 23:22
  • $\begingroup$ Use $\{\}$ for $\{\}$, @exodius. $\endgroup$ – Shaun May 13 '18 at 23:24
  • $\begingroup$ @egreg That just made me think of another question actually if you'd be so kind to answer. If we were to then take the quotient group $\Bbb Z_{120}/\Bbb Z_{60}$, is that group isomorphic to $\Bbb Z_2$? $\endgroup$ – excalibirr May 13 '18 at 23:29
  • $\begingroup$ @Shaun sorry I usually don't forget to write it that way , just made a slip writing that comment :) $\endgroup$ – excalibirr May 13 '18 at 23:29

It's not a trick question! Looks like you made an off-by-one error - it happens to all of us. In particular, you wrote $|M|=11$, presumably because the last value is $11 \times 10$ and you're counting up in $10$s. But let's write out the elements of $M$ fully and see what happens:

$$M= \{0,10,20,30,40,50,60,70,80,90,100,110\}$$ Counting them by hand gives $|M|=12$, and similarly $|N|=30$, as we'd expect for subgroups of $G$.

For a simpler example of this kind of mistake, notice that if we wanted the size of $A = \{0,1,2,3\}$, you might guess it's $3$ because the highest value is $3$, but there are actually four elements - since $0$ is included.

I should also point out that $|M|=12$, $|N|=30$ doesn't imply that $|MN|=360$, see if you can spot why.

  • $\begingroup$ Is it because we should consider MN like a coset and take 0+N, 10+N ,20+N etc ? $\endgroup$ – excalibirr May 13 '18 at 23:02
  • $\begingroup$ Not exactly - look at how $MN$ is defined $\endgroup$ – B. Mehta May 13 '18 at 23:05

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