First I factorize $196=2^2\cdot 7^2$ and therefore the number of distinct Abelian groups is $P(2)\cdot P(2)=4$, where $P(\cdot)$ is the number of partitions of a natural number. So there are $4$ abelian groups. The isomorphic classes are:

The orders are not coprime so they're not cyclic? $$G_1=C_2\times C_2 \times C_7\times C_7$$ $$G_3=C_2\times C_2 \times C_{49}$$ $$G_4=C_4 \times C_7\times C_7$$ The orderes are coprime so this group is cyclic $$G=C_4 \times C_{49}$$

Now I know there is $1$ non-abelian group for sure, namely the dihedral group of the regular $98$ gon, is there anyway to tell how many there are? $$D_{196}=C_2\cdot C_{98}$$ I know the number of total groups is bounded by $n^{n^2}$ and the table with basic facts doesn't include $n=p^2q^2$. Not homework. The table T'm talking about is here https://groupprops.subwiki.org/wiki/Number_of_groups_of_given_order

  • $\begingroup$ If you know dihedral groups, you can use the dihedral groups of order $14$ and $28$ and $98$ in direct products. $\endgroup$ Commented Feb 17, 2018 at 8:26
  • $\begingroup$ If you look on GAP, you find 8 nonabelian groups order 196. $\endgroup$
    – almagest
    Commented Feb 17, 2018 at 19:57

1 Answer 1


Using Sylow's third theorem, there is a normal Sylow $7$-subgroup, so the group is a semidirect product of a group of order $4$ acting on a group of order $49$. You need to classify the actions of $C_2\times C_2$ or $C_4$ on $C_7\times C_7$ or $C_{49}$.

Some interesting examples (not an exhaustive list). $C_4$ acts on $C_7\times C_7$ via the matrix $\pmatrix{0&1\\-1&0}$. This group has presentation $a^4=b^7=c^7=e$, $bc=cb$, $aba^{-1}=c$, $aca^{-1}=b^{-1}$.

Also $C_4$ acts on $C_{49}$ by the generator acting as inversion. Presentation: $a^4=b^{49}=e$, $aba^{-1}=b^{-1}$.

There are others....


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